Okay, so I'm going to continue my series of lectures on eigenvector problems in random matrix theory and estimation problems using eigenvectors. So, today I'll be using quite a bit of, of free probability theory and and transform that I've written there. And so we'll be using them today so it's a bit abstract like that it's just a bunch of formulas. Well, first, I'll start with reviewing those that I've done for it for for addition so, and then I'll re raise it because today I'll be mostly doing multiple matrix multiplication. But in the case of addition, this is much more. Suppose, whoever does free probability or, or, or, or use a free policy in random matrices knows about the additive problem, but just, let me just review it. Suppose that I have a matrix C that's built out of the sum of two free matrices, which here I've expressed the fact that a and B are free by explicitly rotating be the matrix B by a random rotation. If there's a random rotation between a and B, then, then you can use this, but you don't necessarily need this random rotation. If, if either matrix is, is by self rotation invariance like a Wigner matrix or white wishart. Okay, but to make sure. And so, when you have such a, such a matrix addition a free addition, then there's a transform that's extremely useful called the R transform, and the R transform is additive, additive. It behaves a little bit like the generating function of the cumulants in, like the log of the log of the characteristic function in the standard probability theory. So the log of the characteristic function in standard probability theory, give you the cumulants, and the cumulants are additive under normal convolution, normal addition of independent variables. So the key for free non-commutative variables is this R transform that additive, and how is the R transform build. Well, first you build the resolvent which is a random matrix built from the random matrix A. You take the normalized trace for me tau here is always trace divided by the size of the matrix. So it's okay, but, but in informal, informal algebraic setting, you just have an operator. So what I'm going to introduce I think can me forgot what's your notation is fire. It's fire. Yeah, so some people use five some people use tau, I prefer tau because it's closer to the trace. But so my towel is chemist fine. Yeah. And so this is a normalized trace operator, and it converges for Z outside of where the eigenvalues are this actually converges to a deterministic function. Whereas, you should really think as this, this object here is a is an n by n matrix it doesn't converge to anything. It's a, it's a random object. It's a random object with poles at the eigenvalues and and projectors. And the residue is a projector on the, on the eigenvector, but this is a this is a function that converges typically has a branch cut where the eigenvalues sometimes as directs, direct masses show up as poles. So this is a function G, and it's, and since I'm looking at bounded objects, it's always nice and analytic away from the where the eigenvalues are so near near infinity, if things if things are bounded. It never extend all the way to infinity. So there's always a region where it where G is well defined it's monotonic. And so if it's well defined and monotonic then it can be inverted. So kind of a abusive notation but I find very useful. I have G, G of Z, and then it's functional inverse I write Z of G. So Z of G is just a functional inverse of this guy. And it's as a Taylor series at infinity, it always starts with one over G, which I subtract to build the R transform, and then then this becomes regular. It becomes regular at zero. Okay, so this function, the R transform is regular at zero. It's the functional inverse of the steel just transform minus one over G, and another, and this, these relationship. The fact that it's additive, and from the definition you can work it backwards. It's a subordination relation, which tells you that the steel just transform of C, at some point. Z is related to the steel just transform of one of the two matrices. In this case I chose a evaluated the point, which depends on the R transform of B, and the steel just transform itself I'm trying to compute this so this is a this is a nasty equation because it's a implicit equation I have GFC here and GFC there, but it's actually quite useful. And what's very strong is that this is true for these are still just transform so these are functions, this is a function of a complex variable into the complex plane. But this, these relationship extend to these big matrices here that are random. Well, but instead of getting, but, but of course, this is not an equality between random matrices it's an equality between the expectation value of a random matrix. And so here for instance I assume that the matrix a is deterministic. So if the matrix a the deterministic, this is a this is a deterministic function of its argument. And so, if I average, I should say here, if I average over the noise of five over B, or five over the rotations of B, then I have this in relationship that this big, this big matrix G of a, or GFC in this case, so which would be z identity minus C inverse. So if the expectation with respect to the noise and be, then I get a relationship with with G of a which is the identity minus a, but where is the, the argument is the same argument as this guy here. And that's extremely useful to compute, because in, in, in this big matrix, I don't I have eigenvalue information but also have eigenvector information because I have projectors on all the eigenvectors. So I have, and so I have a relationship about average eigenvectors and when I'm going to sandwich this relationship with, with certain vectors that I'm interested in, I can compute overlap. So, how the offer for me that well, when I say overlap I really mean square overlap which is the dot product square of two vectors. Okay, so, but today I'm not going to use this I'm going to use something something slightly more complicated which is a multiplicative case. So there's a whole zoo of, of transform for the multiplicative case. I want to explain right now the multiplicative case, I'll, I'll, I'll explain it when I need it. Okay, so when when I, it's just the one thing that the for the multiplicative case you could you could use the, the still just transform, but the equations are a bit more clumsy. And so there's a transformer that's more natural here is the, what I call the T transform. So, and so, well, the object I'll be looking at is not no longer z identity minus a inverse is the matrix, the product of this times the matrix a. Okay. It turns out that this object of the equation are much simpler. So, I'll try to build. If I try if I can build this combination matrix a, I use data, because I use a different Z because I want to when I take the inverse. I can say that this is the inverse of this function and not the inverse of that function. This is the, the drawback of this notation saying that function. But the point is that I want to build a function which is a times zeta identity minus a inverse. And if I can build this function, then I can say a lot of things about this because of all these powerful relationship that I explained better when when I need it. So the other thing I want to say is that when I talked about the, the problem of phase retrieval. Remember, I define a function H of Z, which was really the normalized trace of H H transpose. And I didn't compute this object and I still won't do it, but I'll show just give you a hint that the technology I'll develop today for this estimation problem can be used to compute this object. Okay, but it will be to take too much time as just a computation. But once you know how to use, if you know how to use this technology, then it's relatively straightforward to compute something like this. So G to two was, was of Z was an object of this type Z one. The identity minus M to two inverse and M to two is really a color was really a color wish hard was really age F H transpose where F is a diagonal matrix of the objects F of Y. And the important thing is that it was F of Y only depends on the element one, which are no longer in the matrix to two. So F is really the diagonal element of F were independent of this this wish are these these white matrices. So basically what I have here is some sort of colored wish art. So a color which are with with the color in the middle rather than the outside, but it's still some sort of colored wish art matrix. And I needed to compute, and I have it's resolvent, and I wanted to compute the normalized trace of a white wish art with the colored version. And if I can use and and the and the and the this G here is will be related to the T transform and have a subordination relation for a T transform, so I can compute all this, and it will depend. In the end it will depend on the S transform I haven't really told you what the S transform of this F. Okay, so, so basically if I know the statistics of F, I can compute its S transform, and I can compute all this. So I won't do this but we will do something similar in this problem of, but I just want to say that problems like that are solvable. That's that's all. And then, well, just advertisement from my book. It's in there. Okay. But now, my point was to go back to to covariance matrices so I talked about covariance matrices on Monday. So today the two sessions will be on this on on covariance matrices. And so really there's, there's a direct problem and there's the inverse problem. So the direct problem is I have a matrix C that I know. And I measure some data age, and I convinced you that you can model the data square root of C H zero. Okay, so H is some rectangular matrix, where the columns have covariance C. Okay, and a zero is a white rectangular matrix. So, so a zero is just a n by T ID and zero one. And then I color it by multiplying by square root of C. And if you if you actually do the random number generation, this is the way to generate correlated data. Okay, you multiply by a matrix on the left, and then all the columns are correlated within the columns and then each column is independent. Okay, so you have independent columns and correlation within the columns in this setting. Okay, so the direct problem and then. I measure a sample covariance matrix as one over T H H transpose, which can be modeled as one over T square root of C, H, H, H not. And this combination, H not H not transpose over T is a white wishart. So I can really write this as square root of C. And I put a Q here, because a wishart depends on the parameter Q and Q is the ratio of N over T. Okay, and remember, standard statistics is when T goes to infinity at M six, so standard statistics is when Q goes to zero. But we're interested in a limit where Q is finite. And I told you that in the direct problem. In the direct problem what happens is that if Q is greater than one, then he has some zeros, I get zero again values, and actually quite a lot of them. Okay, and but also if Q is greater than zero. It's variance. So trace of E squared minus trace of E squared is is equal to the variance of C plus Q. Okay, so this this this this variance here that I computing, you can think of it as a variance of the eigenvalue distribution. The eigenvalue distribution is wider for the sample covariance matrix and from the from the true covariance matrix. It's wider by a factor of Q. So again if Q is zero, then the has the same or if Q is zero, really E converges to C. If it's one zero then he doesn't converge to see you can already see that it's he has the same mean, which I will always choose to be one. So I'll normalize things always that. So, so I'll choose tower C to be one. Okay, so tower C equals one means that the, the, the typical eigenvalue of C is one, the average eigenvalue of C is one. So this is just a normalization I can always scale my problem to have this condition. And so, so see some variance. except, well, see, see could be the identity sees the identity then see has no variance. And then this term is zero. So when sees the identity, the data, the idea that the eigenvalue distribution of the identity. So this is row of lambda. It's just a direct one. Okay, so this is for C equals the identity. And in this case, the variance is zero, but the variance of e is non zero, it's q. Okay, and actually in that case, we know e, e is called the martian capacitor distribution. So, in that case you would have that e as the shape from lambda minus two lambda plus with lambda plus or minus equals one plus or minus q squared. And again, if q goes to zero, then lambda plus equals lambda minus equals one, and the martian capacitor conversion Iraq. Okay, but in general, for q non zero, but this is true for q between. Well, this is, I won't discuss much the q that the case q greater than one, but in the case q greater than one, you also have a dirac at zero. Okay, but but the the martian capacitor. At least the field just transformed the martian capacitor has a pole at zero precisely when this happened but anyway, I won't have time so let's forget about the case q equals one let's just assume now that q is between zero and one. But the case q greater than one can be solved is just so they're not too, too complex if I too much and they're going to talk about. Okay, so. And I found, and the other thing is that I, I showed the equation for a slightly very different version of this. I showed that that there's a there's a there's an integral equation. If you if you write E, not like this one, but a similar problem. If you write this problem. Okay. Then by considering E as a sum of projectors. You can write an integral equation related to the eigenvalues of C to this, the eigenvalues, the spectrum of E, and this was actually the work of martian coin pastore and 67. But by by by moving around matrices and considering matrices that have similar eigenvalues you can actually find a spectrum the same method for this problem is just a bit more cumbersome. There's a more direct method to compute this, and it's using these transforms. Okay, so this is again, still in the direct problem. And now can erase. So actually today we won't be doing the additive case at all. I guess can read but the additive. I mean, if you, if you need to know one transform should know the are the s, the s you should only know you can forget about it. And, and relearn about it when you need to do multiplicative problems, but the are you should know by heart mean the R is super useful. It's everywhere. Okay, and there are even relationship between the R and the S. But anyway, but today we're going to talk about the S. So maybe it's a good time now to. So you see. Okay, the notation is bad because it's on this. So my sample covariance matrix. So maybe you just write this is here. This is important. So my sample covariance matrix is the square root of a matrix. You multiply and the, well, this is also just it happens naturally but you want it's important that the matrix that is symmetrical. So, or if you can do all this with with with her mission matrices, and in the complex case, and it's this case is would be an Hermitian matrix. So it has real eigenvalues, and it can be diagonalized. So it's nice to work with the symmetric product. So if I go back here, I really want to do the product of the matrix also doesn't. So here, really morally what this is the product of the matrix a and the matrix B. So I'm assuming that a and B are positive definite. Okay, I mean I can relax these assumptions but for the will be considering they could have zero eigenvalues and one of them could be non positive definite but let's just assume it's a simpler to assume that both matrices are positive. They are positive definite. So I can take the square root, the square root of the matrix I recall you, you can, you can there are many ways of seeing it but that's the one that's that that's symmetrical is you, you diagonalize the matrix.  Okay, so, so by square root of the matrix. So if I have a which is equal to all D transpose square root of a is really defined as all square root of D transpose, and square root of a diagonal matrix is really square root of the elements on the diagonal. So this, and, and I want to put a square root of a on both sides, sort of a symmetric matrix. Okay. This is kind of object and I'm, and I mentioned this matrix Oh here is only really needed to ensure that the problem is rotation invariant here in the middle of a white wishart.  For a white if B is a white wishart I don't need these matrices, I have I do have a free product. Okay, so what I say is that in this problem there exists a transform S, which is multiplicative so the transform as the S transform of this matrix C will be equal to the S transform matrix a and the S transform of the matrix B. Okay, and Of course, of course, all this is a synthetic. Yeah, yeah, yeah, yeah, it's only true. Yeah, so basically, at finite and they're one over and corrections all this. Okay, so, so yeah, of course, yeah, so, yeah, so this notion of freeness doesn't apply. So basically you. So, large, rotationally invariant matrices are free up to factor one over N. So, if you look at the equation that you need to satisfy to actually be mathematically free. You see that there's a one their terms in one over and, and so it's only that's why we say that large that random matrices are asymptotically free, they're free. They, when the size goes to infinity, but for us we're going to be talking about matrices of size 1000 in real applications and one over 1000 small enough given all the other modeling error that we make that this is not the main error. Okay. So, so, Yeah, so the, I have these are these S transformer multiply and the definition of the S transform is similar to the R which I've, so you need you have a you have a transform on the matrix itself in this case you're just kind of the equivalent of a we call maybe a T resolvent or in a T matrix. And then the normalized rate of this T matrix is some function. So it converges in a complex plane outside of where the eigenvalues are, it's actually related to the steel just transform. So it's just, it's just zeta g of zeta minus one. So it's, it's, it's very nicely analytic outside of where it is very similar has it has a branch cut where the eigenvalues it can have poles. So the eigenvalues it's, it's analytical, and, and it's a regular at infinity. This is just that the behavior at infinity slightly different, is that the, the exponent of one over Z are shifted by one, and you don't have the one over Z pole at infinity. Okay. It's actually new overseas but it doesn't matter. Anyway, so from this, you can compute the S transform, which has this for some reason. It's T plus one over T in an inverse of this T transform. And more, more beautifully, you have these, these subordination relation, so the T transform evaluated at one point of the product matrix is the T transform of one of the matrices evaluated at the different point that depends on the S transform of the other. And what, what I'll be using for to compute eigenvectors is this subordination relation. So if you think of this T matrix T matrix again, like the like the resolvent it's a random matrix but I can take its expectation value with respect to be. So typically I'll take the expectation value with respect to the, to the white wishart, which we can think of it as a random matrix. So if I have if, if a is a fixed matrix, and B is a white wishart for instance, then I can average over the white wishart. And then I get the relationship that the average of, of this object for for the product matrix is given by this object for the determinants a evaluate at some point to the same point as here. And this point depends on the S transform. And the last thing I want to say is that why this is also useful is because white wishart have a very simple S transform. So if I have a white wishart parameter q s q of T is very simple is one over one plus QT. Okay, so, so if I'm averaging over a white wishart this term is relatively simple here. Okay, in general, it can be quite nasty. This S transform. Yeah, the subordination relation on the on the T matrices here that this generalized the resolvent, does it become true without the average in the large size limit. No, no, because this is a, this is a matrix with the eigenvectors are all over the place. It's an n square object. It doesn't. Yeah, but I mean the resolvent for example, can this one yes. Yeah, this is this is a scalar equation. No, no, but the resolvent seen as a matrix that there are all these notions of local laws and they, and in certain scales I mean you have concentrations at the level of the element, but I'm not too close to the real axis now. Because this this is a this is an object with n square entries. And I don't see how all of them know that the eigenvectors fluctuate. So, for instance, I think this way if, if a is diagonal, if a is diagonal T of a is also diagonal. And then without the average, it would mean that T of C is diagonal. And it just cannot be it's a random matrix, it will have off diagonal entries, is it that on average the off diagonal entries are zero, but I think they may tend to zero in the large size limit. And for the resolvent of a being there, I think that's the case. We can discuss some I really don't think so. I think that that you would get absurdity to actually use that in some of your papers. Yeah, you can get generalized elements that you can do it for any vector and so you can take a unit vector with just one component what you want. And this shows that each element essentially concentrates if you're not too close to the. Okay, yeah. Okay, yeah, but it depends how you use it, because yeah. Okay, so I see what you mean. But it with this kind of argument, you say that a bigger matrix is a zero matrix. Because, because all the elements are small in a bigger matrix, all the elements are small. They all, if you look at one element of a bigger matrix, it converges to zero. And but the whole matrix itself when you multiply it, you multiply and and such random matrices, such a random elements. And the fact that you have any of them. They're all small, but it becomes big. So, so indeed, okay, this is quite subtle, and thank you for pointing that out. Indeed. But I think this is the best way to see it a bigger matrix is essentially element wise, the zero matrix, but collectively, it's definitely not the zero matrix. Yeah, so you can, yeah, so it's okay that this. So, indeed, a point wise if you only look zoom on one element, and you're not going to use n elements, then, then you're right. Yes. Okay, now I see what you're trying to say. But you have to be careful with these things, you can't treat them collectively anymore. If you say one element converges to zero. And then such elements don't, it's very dangerous to treat n elements, because because they convert to zero as one of our square root of n, or was one of ran. And so when you have any of them, then, then you then you get a spectrum of eigenvalue, this is the bigger matrix definitely not the zero matrix it has eigenvalues that are between minus two and two. The zero matrix has all eigenvalues equal to zero. So, so, and of course, the, the eigenvalues as a function of all the elements. So function, so you can each element convert to zero, but all the elements can treat it together, they're definitely not non zero. So, okay, so if to be careful. Okay, I mean, being a physicist you have to, to, to be careful what you do where we're going to addition you make precise statements, and, and, and vague statements need to be, you need to be careful with vague statements. Okay, anyway, so, I'm going to rewrite. So basically, if I have E equals square root of C, we should q square root of C, then the subordination relation tells me that T E of zeta equals T C of zeta. And I know also that the S of a white wishart of parameter q is one over one plus qt. So I have a subordination relation which I can write as zeta one plus q T C of theta. And this is equivalent to the ugly integral equation that Martin point best or wrote. Okay, so if you know C. You can compute values of C, you can compute this it's T transform, and this is an this is an implicit equation because T E here with T E. Yeah. Okay, so, so it's an implicit equation because I have T E here, but I also have T here. In general, it's pretty ugly. This one simple case where C is the identity. So the T matrix is identity, zeta identity minus identity inverse. And this is just essentially the matrix one over zeta minus one. It's really matrix. Okay, maybe I'll write it this way. It's one of zeta minus one times the identity. Okay, so the T matrix of the identity is super easy to compute and therefore the T transform. So the T of the identity is just one over zeta minus one. And then I won't do it. But if I say that see here as this very simple T transform. I can develop all this, I get a quadratic equation in T. And if I solve this quadratic equation in T, I get the T transform of the marching purpose to her law. And with the relationship between the steel just and the, and the T transform, I can recover this this there. I can I can also recover the density. I can also you can also there, you can also recover directly the density. I mean, the from the T transform there's a branch cut, and the branch cut is instead of just being the density is lambda times the density. So, so the T, so basically here what I'm saying, if I do this in this, this very simple case, the white wish art. Then I get a quadratic equation, and there's a branch cut offer for T that I get. And on the branch cut I can recover the the Martian Cooper store law. Okay, so this is a very powerful way of recovering Martian Cooper store, but it is what it's very powerful because I can also do any colored one. There's one case also that's solvable. That you know, you should know about. I'll talk more about it. Is what it's called the inverse which art. It's one of the few. Well, they're basically only two, two random matrix that are very simple. They're there, not maybe, but they're, they're as transforms are typically very hard to compute, but they're few simple case one of them is the wish art, and the inverse which art, which would, which is the inverse of a wish art matrix where I rescaled it to have mean one and variance P, it would have S of P equals to one plus PT. So in another case of a very simple. So, the inverse we started to take a wish art matrix, you invert it. Well, the inverse doesn't have quite mean one so you change the mean, it has some some some variance which depends on q, which you rename P. So this is a, this is a matrix with mean one and variance P, but it's really built from from the inverse of a wish art matrix. And this object has a very simple S transform, which means that if I injected here. I also get a quadratic equation. That's a bit more nasty because it's a quadratic equation that will depend on P and q, but I can do and I can have an explicit quadratic formula and basically the only thing we can do. In life or quadratic functions anything cubic or Fort order is too messy to, to, to, to work with, but for this case you get a quadratic equation. And, and this is going to be very important, because so the Jacobi ensemble of random matrices is essentially the product of a wish art and inverse which art. That's one one application. The other one is in Bayesian inference. If you if you're trying to, if you this was this this problem that I'm going to talk about inferring the matrix C was done by statisticians. I think in around 1980 or late 70s. And they say well the only thing you could do is an inverse wish art, because that's the conjugate prior. So if you know about Bayesian inference. So if you look at the prior that's the easiest to work with. It turns out that inverse wish art are is a, and you can do that in finite dimension. I mean there's a Pierre Palo showed you the, the wish or distribution for finite and the inverse wish art is also known for finite and and for finite and their conjugate in the in the Bayesian sense. So if you have a Bayesian theory of random matrices, then the easiest prior you can work with is inverse wish art, and you can do this inference problem and if you assume an inverse wish art, you get very simple answers. Okay. If, if, if your matrix C is either white or inverse wish art, you get simple answers. Okay, in other cases, you have to do things numerically. Okay. Yeah, let me just start. Let me state before we break the inference problem that I'm going to talk about. Okay, so So basically, I have a matrix C that I don't know, I measure, but, but I measure. E, which is, I model a square root of C white wish art square root of C, but I can model with any multiple, I could actually So in theory, I'm going to develop is a bit more general than that. I can say, I measure, I have a matrix C that I don't know, and I have multiple some some multiplicative noise. Just going to call this w, but w is not necessarily wish art matrix is some noisy matrix that is one on average, but could be and in some from financial application. It's just that that you have all sorts of correlation here we're talking about the, you can have fluctuating variants for instance the fact that volatility fluctuates. There are all sorts of noise that can happen that most of the kids doing that noise that could happen here. And it's just as easy to do theory, not just for a white wish art but for any multiplicative noise. Okay, so my problem. So the two things that I want is that. So, on average, my noise is one, and that might that w is rotation invariant. So so I can be so I can use. So this is really the free product. So really, he is in a sense that is a is a is C is a symmetric free product with with with C on the with C on the outside. Okay. So, the question is how I want to build an estimator. All right, Kai. Kaiser. So we can be a. So I observe E so it has to be a function of E. And I want Kai of E to be an estimator of C. So I don't know C. I don't know what it was generated by C, but I don't know it and I observe E so I want to build an estimator from C. Okay, and the problem is rotation invariant the noise is rotation invariant. And I don't know what the noise is rotation invariant and the matrix C itself. I don't know what it is. So I could do it sort of a Bayesian approach. I would need to make to build a prior, and I'm saying is that I have no prior for the eigenvectors. Okay. Of of of C of the truth see another way of saying that is that since the problem is completely rotation invariant I'm going to look in the class of estimators that are rotation invariant what's a rotation invariant. This is an important concept as a rotation E invariant estimator is a. So if I, if I'm given a rotated matrix E, then I'm going to have the same estimate as a non rotated E up to a rotation. So, an estimator that has this property. Okay, and what I argue that since if you have no prior on C, then, then necessarily your estimator should be in this class. Another way, and then the consequence of this. Okay. And maybe you'll have more intuition on the consequence is that your estimator E commutes with E. In another way, another way, kind of E is in the same is in the same basis. Okay, so if you have an estimator in this class, or you can think directly as maybe it's easier to build the intuition about about this statement is that you don't know anything my eigenvectors, you're trying to estimate the matrix see but you really have no prior knowledge of its eigenvectors, and the only eigenvectors you have in the problems are those that you observe those of E.  Other than the eigenvectors of E. So which means that your estimator should stay in the eigenvectors of E, and this is a very strong. basis. And of course, like, to example where this fails. If you think that he might be the C might be sparse. So if you have a canonical basis. So, so if, well, first of all, if you have a canonical basis if you truly believe that there exists a basis in which C is natural. And then C might be diagonal on that basis and this is this breaks rotation variance. You might have a collective modes is your, your, your, like in finance for instance, there's a, there's a, there's a canonical basis which is every stock is stock one stock two stock three that that that's a basis. There's there's actually an eigen mode that's very easy to call the market mode where all the stocks go up or all the stocks go down together, and that breaks rotation variance. Okay, and the other type of thing that can break rotation variance is the notion of sparsity. So if you say that that the eigenvectors of C are sparse, that also breaks rotation variance. So, so this part is very strong. And, and in some cases not justified. But, but, but on the other hand it's very powerful, because essentially, and I'm going to stop here. You go from an n square problem you're trying to estimate and the n square elements of the matrix C. Okay, but you really only need to me just write finally that Kai of E, you can write it as sum over K. Psi k, vk, vk transpose, and just to set the notation, E I diagonalize as sum over K is lambda k, vk, vk transpose. Okay, so if I diagonalize the matrix E, and the V, the V case or its eigenvector, I'm saying that my estimator must be in the same basis. But, but the fact that it's in the same basis doesn't say anything about the coefficient. Okay, so basically I'm I reduce my problem from an square problem to estimating and coefficient and estimating and coefficient is much easier than estimating n square matrix entries. So this rotation variance hypothesis is extremely strong. It's strong in two way it reduces that radically the national table problem, and it allows you to use very powerful tool of, of rotation invariant matrices which, which, when they're large become asymptotically free. And then you can use all the work of the school on on on freedom. Okay, so I'm going to break here and reconvene unless there are questions about this. Is it so clear that this class estimator in, I mean, apart from the natural argument that, yeah, you don't have prior and so the only basis you know is the one of the data. Is there like stronger argument that shows that it is like the Bayesian optimal that belongs to that class. Yes, if you if you write a Bayesian estimation, and you say that your prior on C is rotation invariant. Then you can show, you can show this property. Yeah. So, so yeah, that's what that's to me that from, I like Bayesian estimation so I am comfortable with it. So in that sense, to me that's the best proof of this is that you write P of C given E as P of E given C times a prior P of C, you say the prior P of C is rotation invariant you look at that little terms in equation, and you can show that you can show first that that the estimator satisfies this property. And, and, and you can easily convince yourself that it will commute with the matrix E and it says actually with, I won't do this because it's almost it's too abstract that it's super quick to do it really is kind of kind in a, in a, in a free. If you forget about random matrices and you just think of these as non commuting objects, and you can actually write that this estimator is just going to be the conditional expectation of, of C given E. If you write this abstractly, you get very quickly the little apache equations, but I think this is a bit too abstract. I have a little section in the book where I do that but, but you don't get any, any insight on where it comes from. And then then in free. Well, I think, yeah. Again, I'm not an expert on free probability but but for for people in free policies it's pretty obvious that the conditional expectation of E of C given E must commute with E, but that's a bit abstract to me so yeah. Yes, can can do. So this can be the hypothesis about rotation variance can be relaxed to permutation and variance. I don't know how to solve the permutation there and probably been indeed permutation variance is much more natural. But, but, but much harder to. But can you do can you do the inference from Okay, unfortunately, unfortunately, okay. Okay, so because I'd be really curious of the permutation and variance prior. Okay, so that. Okay. Yeah, so I don't know but I don't know if in the end you get, you can get a practical answer that you can build a portfolio and what do you tomorrow send me the answer that I won't be there unfortunately but okay so let's break maybe.


You're recording, right? No. Why, what are you gonna do? Nothing, I just need to know if I need to protect my nipples and not take a drink. Always protect your nipples. Always, kids. Hey, welcome back to where Stephen rakes to be at. It's I'm Corbin. Words of wisdom from Corbin and Rick. It's so juicy. It's so juicy. Oh, no. Follow us on our personal YouTube channels, links are down there.  We're both wearing blue. Krishna. Krishna, I don't know. The god? Is he blue? Is he blue? Blue is like his color. Yeah. I understand that. Today we're reacting to a top 10 places in India you are not allowed to visit. Well, right now we can't go to any place in India. That's true. We can't go to any really moot point. So that would be number one on the list. That's number one on the list, the whole country. You cannot visit there. There are places you shouldn't be able to visit. All those roads that we saw. Yeah, that you have to drive. I got messages. I'm sure you did too. Yeah. Hats off to all of you who have done that. Great stories. I've read some stories that are pretty amazing. Obviously when COVID is over and the restrictions are gone, which is starting to ease some things up actually in some places. Like CDC has said, you don't need to do a 14 day quarantine anymore in some places. And here in America, we're just brimming with COVID fund. Well, India is as well. Yeah, it is. America, India and Brazil right now. Yes. Winning! Anyways, but America is number one. Anyways, I don't know what I'm saying. Here we go. Top 10 places you're not allowed to visit in India. Okay. Hello friends, welcome to our YouTube channel. India is the largest country in South Asia. It is famous for its many traditions. But do you know that there are many places where nobody knows about them? Come, let's take a look at them. Number one, North Sentinel. North Sentinel Island, the part of the Andaman Group of Islands. Oh yeah, I know that. They're crazy men. They're crazy men. I remember that shot. Okay, just leave him alone. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. If you want to set foot in Bangar, then it will be illegal against you if you get out of the sun. Bangar is one of the most dangerous and most dangerous places in the world during the day. You should definitely leave before it gets dark. Number 4. Number 4. Anyway, this is the second place where Andaman and Nicobar are part of this list. And for good reason, the population of Andaman, Nicobar Island, are really small. UNESCO has presented the UNESCO World Heritage Network as part of the Biosphere Reserve. It is a platform for wildlife, marine animals, and plants. It is a space for these areas. And it is a space for civilization. This is the reason why Nicobar comes here. That is why the population of Andaman is at least less. It is very difficult to get a permit to go to Indivipo. And this is mostly due to scientific and research. It is better this way because it is necessary to keep the population of Indivipo ancient and unknown. Number 5.  It is forbidden to go to common people here. Because this is a military area. And there are also terrorist organizations here. People can't even think about going here. Number 6. It was very beautiful. Some of India's most famous dacoits have been gone for a long time. This long time Bollywood film was going to portray the art of civilization. However, there is still a lot of danger to the victims of the Shesliya Decades. And that is why it is safe to visit here. Wow! Number 7. Bhasar Chattisgarh. A district of Chitrakot Bhasar Chattisgarh, Nehagra Falka Garh. In which there are many nationalities. The whole village is compared to the natural beauty of this place. This red-collar is one of the first places to fall from the Durbhag. Number 8. This is another place in India where it is said that common people are respected. Some people say that a special permit can be taken here. But according to our research, this permit is only for the army. The reason for this is that this place is very close to the border of China. Anyway, the people who say that this is one of the most beautiful rivers in the world. Number 9. You can see the red-hot lava.  However, no one is allowed to go to the mountains for safety reasons. Number 10. Obviously. I knew that last one expected that one actually. Yeah, that one obviously has a lot of going-ons right now. I've never thought about it. How many volcanoes does India have? I've got mountain ranges, so it makes sense that they would have some places that have volcanoes. I'm sure the fault line is probably in the Himalayas, you think? It's got to be somewhere around there. Yeah, so I wonder how many volcanoes you've got. We've only technically one in America. Well, we've got a little dormant. St. Helens is still active in Washington. Washington and obviously Hawaii. I don't know if Alaska has any. I don't know if they're active. The Ring of Fire, which lasts all the way from America down to California, all the way over to Japan. Part of the Pacific Plate. Which is why it's one of the most active. Yeah, the plate moves over a fissure. That's where you get for the Hawaiian Islands is an active fissure. So as the plate moves, that's where the Hawaiian Islands keep getting birthed and get more and more. Yeah, it's very cool. I didn't know. I've known about that with the indigenous people. The indigenous people, I didn't know that was India. I thought it was closer down to what Papua New Guinea and the Pacific Islands. But hey, it's definitely a famous island. It definitely makes you curious. I just leave them alone. They seem pretty one. Obviously, it wouldn't be hard if you wanted to take over the island because I'm pretty sure they're pretty native. But why? You don't want to disturb them. We don't want them in this world. No, and any places where that's happened, the indigenous peoples, they typically don't respond well to that. No, at all. They're in a different world than we are. Papua New Guinea, for the most part, Papua New Guinea has now been touched. But for a very long time and not too long ago, it was one of the more primitive places that had untouched people groups in Papua New Guinea. And I love the fact that the one island is just untouched just to be a haven for nature. And the animal is kind of like Madagascar. And I wish there was more of that. Me too.  Too bad we don't know how to steward things. No, because we're just stupiding things. We're stupiding things. We're stupiding things. Just like that sentence. But that was cool. Let us know more videos such as this or ones that will blow our mind like the road ones. Scarce. Yeah. You guys like seeing us freak out. Yeah, but see, you learned something new. I didn't know that that place was there and I didn't know about all those other places. So we like learning because that makes us less stupider. Sometimes maybe seven.


Hello everybody Welcome to DCU Silver Sock presents mastermind. I'm your host Shaulk Harry, and I'm joined by the chair of Silver Sock mr Andrew Dennis Cleason. Thank you Sean. How are you Andrew? I'm pretty good You know what? We're ready for today ready for it to see people's knowledge And we certainly have got a lot of knowledge a lot of knowledge today And as a wise man once said mr. Benjamin Franklin an investment in knowledge pays the best interest That's what he said. Yeah, so Yeah, so we have a few contestants From different societies today. Give me one of them. You got that right Sean Up first we're gonna have Podrick Henry Podrick Henry Podge New man who is doing something with welfare and something like that for the council And he is going to talk to us about superheroes Then we also have Aiden Lawless Mr. DJ sock man He's gonna tell us about music in the house version of it and then then we have Mr. Ocean O'Connell Mr. Ocean O'Connell is gonna talk all about his favorite book Harry Potter Being a long-time Harry Potter fan.  I can't wait to see what plays out tonight Andy. I'm really excited So without further ado should we welcome our first contestant to this to the board? I think we should Aiden Lawless, please get yourself up here my man Thank you studio audience Nice to meet you. It's great to see you. It's great to see you. You're looking man. And so listen Aiden What do you know about house music? Do you like it a little bit? Yeah, I suppose you can say that cool Cool, that's pretty rad. So we have a few questions for you Aiden The way it's gonna work guys today is we're gonna start counting up every every correct answer the person gets And it's out of 20 questions the best out of 20 wins a prize the members no likey no likey So let's rock and roll with the first question. You ready?  Osef. Yes, that's correct my friend score one for Aiden name all three former members of Swedish house mafia Oh, no I don't got a pen. No, it's okay guys cross is working Steven Grosso. Yeah Well, yeah one more Yeah, so far question number three Jax Jones track, you don't know me is huge at the moment. What is the original? Song song what is the original song? What's it song? What's it sampled from? Yeah? Oh Is it book of shades? It's book of shade anyway Do you know what we'll give him a point for that now I think you deserve Body language. Seriously guys anyone got a pan that works. We've got a pan that works anyone Do you got one call? Thank you. Our liaison officer is gonna get us a pen while we continue on with the proceedings And name five tracks by Eric Prydes.  Oh sad man. Aiden. He's still on two points. You're on two points name the duo behind duck sauce behind the group duck sauce Barbarous. Oh, duck sauce. I'm a van helden. Yeah Pass. Oh, no, it's a track. These are some tough questions. I should know kind of gave you I must say What year did Michael Gray release the weekend Nope, there's not 2007 2008 or 2005 No Okay. Oh, no, what is deadmail? It's real name Joel Zimmerman was the answer what ages fat boys slim a 47 B 51 or C 53 47. No, he's 51 sadly Okay, 83 51 51. Sorry. Yeah, I realized that the yeah sunglasses and sunglasses meant the correct answer and listen Aiden relax You're doing you're doing well so far in the lead still in the lead Question number nine back to Eric Prydes. What is Eric Prydes reason for not doing gigs out of America? Why does he never leave the States? It's a rare one Doesn't like airplanes. Yeah, he's afraid of flying. That's pretty good. Here's a here's one for your DJ knowledge What is the exact BPM of Calvin Harris acceptable in the 80s? That's a toughie No, just one option here Okay, and a is it 115 B is it 120 or C is it 125? All unusual BPMs, but you say much I saw myself show. Yeah, and I know a lot about BPM 120 no it was 115 my friend. Okay, the next questions are from jazz. So they might be a little bit easier We're gonna try and see and so yeah Well, listen Yeah, yeah, you're on two points three points three points Frankie Frankie Nichols was an iconic figure in the house music scene. What was the unofficial title that people called him? Godfather has yes, that is one point Next question where was house music said to originate Chicago? Oh, correct in 2016 Patrick topping remixed a track called Damn a Pree, but who was the original artist? Jamie Jones. No, it was Rome Acoustic for Jacks. Oh Jacks Jones popular hit. Oh, we doubled up on questions here. So we did wish we did so And Jacks Jones popular hit you don't know me it's been tough in the charts recently But what is the name this track song samples? So you got it. Yes. That is a point. That's already. Yeah. Yeah He's doing pretty bad test pilot is an ad is of which famous producer Test pilot. Yeah Richie horn dead mouse In whose boiler room set to the pillow fight break out cow cocks No, it was scream and disclosure Yeah, and Simon Neil aka shadow child has another alias. He produces under what is that alias Dave Spoon Name one track name one of the final 10 tracks cow cocks played in his last ever set of his 15 year residency in space Ibiza Detron photo Yes, that's it. That's an answer. Correct. Give him a point And faithless hit insomnia is deemed an absolute classic But what is next in line insomnia? Please release me and let me dream of making mad love to my girl on the Heath Tear enough tights with my teeth. Correct my friend.  You're on eight What is the name of the glass? Weegean DJ jackmasters label? Oh Oh The answer is numbers listen Aiden your final score is eight points. Congratulations. That's almost 30% of the answer There we have it Aiden from DJ sock Very sweaty hands. I must say and is quite the boy when it comes to house music And listen he did well. Will he retain this lead? Yeah, the competition. So he's currently in the lead Let's bring up our next contestant mr. Podge Podger Kenry How are you mr. Vice president good to see you good to see you. How does it feel to be mr. Equality and welfare? Okay What are we gonna ask him about superheroes? Well, we're gonna ask Podge about superheroes today Yeah, so what's your favorite superhero? It's my favorite part of your flash It's a strange answer and we'll say minus one point for parts. I'm only messing guys We don't really do that here in mastermind. What powers would you have if you were a superhero part from obviously welfare? I don't I like the idea of super speed. That's my fashion. I don't know maybe that or Yeah, they're super speed super speed pretty cool. So flash again. Um, okay, but let's get on with the questions Podge, how's that sound? Yeah question number one. What is Superman's weakness? Correct well done Got them is Batman City, but what is the name of the neighborhood neighboring insane asylum? Yeah What is Iron Man's real name? These are pretty easy questions. I could get these how many Captain America films have there been and Can you name them all? there's Captain America The First one the name there's Captain America with soldier and Captain America to the war. That's it. That's the three wow Who founded and was the original leader of the Avengers? Captain America no Nick Fury. Oh, what was spider-man's uncle's first name? Ben yeah Spoiler who died at the end of Batman versus Superman down of justice Yes Range the movie for me. Where's Torcom? I haven't seen that yet I've got it. I've got it. Yes, correct. What newspaper did Peter Parker work for? This is a good one. This is a tuffy. This is a real mind master Bass daily bugle bugle bugle Who played the Joker in the films before Heath Ledger in the motion pictures? Oh Yes, I'm gonna argue the better Joker, but I wouldn't yeah I don't know why I said that because I absolutely isn't I think he is I like I don't think I love Jack Nixon But I don't think he's quiet and there's a time in a place for a sandy. You got to get back on with the show Who what was Iron Man's girlfriend's name? Oh, yeah What is Torah's brother's name? Loki Loki? Well, that's what pot says to all the women sometimes pot is just taking the lead after he kisses them. He says Loki Keep that low key Name four of the five men who have played Batman in the motion pictures. Oh Christian I Won't get your point. Yeah, I know that don't stir the tea Listen, you can't even get too fast You could have got Michael Keaton Val Kilmer George Clooney or Ben Affleck What's the name of Batman's butler? Yeah, how many actors have played spider-man? Three Spoiler alert he has passed by Aidan we do have a new clear victory at the moment and Can you name both of spider-man's girlfriends from the motion pictures? Merging yeah, that interest can you name the actresses? The bonus question Mary Jane something us at sober stocks stay well away from Who is Batman's detective friend? Jim Gordon, yeah What actor plays Magneto in X-Men hint same actor as Gandalf and Lord of the Rings. Oh, um, I can't even I keep in the proper store, but that's Professor X No pass in the Kellan real old dude two more questions, Podge Second-last question. What Avenger has played two Marvel roles. Oh Captain America. Yeah, what was the other role he played? Fantastic four yeah, which character Johnny Close podge. I'm gonna let you take one more That's that second name Human torch. That's okay. Yeah, well you want to have torches and fairness final in the movie the Incredibles What is the name of the family friend whose wife hid his super suit? Where is my super? Where where is my super suit? That's super suit You never bring me out anymore Do you know the answer I said or no no Pretty cool guy well podge. What's the total podge take the lead with a whopping 15 questions right? My friend put it right there. Congratulations. Thank you. I don't say much to you in the final. Okay, so stay around we'll see you next time Okay, so that was podge. He obviously loves his superhero films. Yeah Next up guys, we're gonna have our only duo in the competition because it's such an unfair thing, but This is a duo who thinks they're dynamic, but not really nice today. It's Tyrone Chloe.  So I prepared I prepared all these questions myself because I know all the answers So you can play along at home kids and see how many you get Question number one. Are you ready? Yeah, where is Sean Kerry from? Bill Griggin. It's an easy first point. How many brothers to Sean Kerry have two. Can you name them? Ronan It's pretty impressive Does Sean Kerry have any pets? No, correct, man, you said you're on hide How does Sean Kerry make his money? He works as a radio presenter No, I work as a team here who is Sean Kerry's favorite hip-hop artist Kyle yeah, and What Kanye West obviously What's Sean Kerry's mom's name? correct You guys Google that How did you how did you know that he didn't know that a few hours ago? Cheaters don't give him a point Take a point away. Um, how many relationships has Sean Kerry ever been in one? What is Sean Kerry's oh, I'm gonna take this one out. Yeah I'm gonna make it a little bit harder. Yeah, I'd agree.  Oh, I know this one. Oh, man Yeah Oh Reeboks The answer is new balance girls It was new balance. That's not even man. What song would be the theme tune to Sean Kerry's life if he had the choice? American boy. No, I'm afraid the answer is dying by Mac Miller and Anderson pack Sean Kerry presents a sports show every Saturday morning on f4q.fm But wow, who does he host it with? Nathan Yeah, correct Sean Kerry will appear in the upcoming DCU drama production of breakfast club, but what role will he be playing the principal? What's the name of the name of the principal? I can't do that girls You're on a total of seven if it was up to me a point is up to me I'd give you the point but and he's the point man. He's the point man No, no point man. No point man in here Sean Kerry used to play a sport for four years But what was the sport? and the boxing time No It was cricket Never told us that ever. What's your martial arts? What's your martial arts to Sean training? No Muay Thai and Jiu Jitsu What is Sean Kerry's amateur Muay Thai record?  It's a real wild card.  Yeah, what age was Sean Kerry when he first shifted somewhere? 12 Jesus 14 What what football team does Sean support? Oh my god Do you guys even know me? Do you skip the yeah? I'm gonna go back to it. Okay. It's gonna be a good last question. Oh My god, I always wear the jersey always well, I wear it to the gym sometimes I know is Chelsea. No, it's not Chelsea. It's Everton Okay Who's Sean Kerry's favorite WWE wrestler of all time? That's mine And the last question what is Sean Kerry's name? There you guys got it so after all that what is their cumulative matter of nine, huh triple H With a total of nine points, I'm afraid you have not broken the lead But thank you so much for your time guys. Thank you guys Before we bring on the next the next contest and we're just gonna have a quick run through of how Competition has gone down so far. Yeah, so Sean. Well, listen, we've had we had first We have we had Aiden Aiden was doing house music for us and he got a he got eight points Which was admirable because the questions were bloody hard really impressive the hardest questions the hardest questions. I've ever seen and With some easier questions was mr. Podge Henry and he he got a whopping 15 points What you really sailed through those questions.  Yeah, good promise all season on I really like how he said to develop his game What about the open-comers the debutants? I'm Chloe Tyron Chloe through injury. They made a they made their Appearance it's meant to be Valakio. They got off. They got off to a good start Sean But towards the end. I think they got a bit too cocky and a bit the bit shake They got way too big for their boots. Remember, they're just the backups. They ain't the starters. Okay? and well said Sean so before we move on with the rest of our Procedures, you got to tell you guys that on Thursday at 2 p.m.  Have you seen the video Andy? I've seen the video Sean It's really good. I really enjoyed it. I thought it was really funny and made by the one of us Connell who's our next contestant ocean come on down Put it there my man. How are you mr. MPS? Good to see you. Yeah, take a seat.   Do you like Harry Potter? Oh, yeah. Have you watched all the films? Yes, have you read all the books? Yeah, of course not. He doesn't read What a who's your favorite Harry Potter character? Or Ron Reevesley or Hermione I like I like to move or they's pretty cool. Yeah, I'll do it again. Okay, but are you ready to rock on with the procedures? I'm ready to go. That's number one. What house did Voldemort use as his headquarters? And mouth on money. Whoa, correct. Oh, my god. Question number two Professor Rampus Lupin was affected by Remus Remus He was affected He was affected by that again show like and floppy. What did this do to him? Hey, he turns into a werewolf What platform does the Hogwarts Express depart from none is your Boris? Oh sugar. You're pretty good at this I know what is another name for Hogwarts Lake? The black like that's that's four in a row What do Hermione's parents do for living? They're dentists. Oh, man Who does Guinea go to the Yule Bob? Oh She goes with Neville long-awaited. Yes, she does for you is hot Yeah, now he's hot. Yeah. What is the name of the dragon that Hagrid had?  Where's the last horcrux crooks? It's it's in Harry Like it was reading the script here. Um, what address did Harry live with Vernon and Petunia Dursley? Privetry, but I don't know what number you don't know a number. I'm afraid we have to get the number for she I Think it's four. Oh my god. It is it is for okay Dumbledore has a scar above his left knee That is a perfect map of what? Okay, only if you get three more questions correct you tie with Podge Okay, wow This is a big so what okay? What is the name of the book Hermione gives to Harry before his first-ever quiddish match? Oh god. Oh It's something to do Quiddish through through a through the ages.  Yeah, did you pick this is supposed to be for me? Wasn't it? Yeah, I didn't take that I didn't he wrote the questions. I didn't write the questions. Oh, sorry What is the name of the three-headed dog in the first book? Fluffy, correct Who is the leader of the centaurs? Oh? I have an abrasion the answer is wrong. Yeah, machine. Okay. That is bad for him Two magical skills other than Hogwarts competed in the twizor tournament Can't read name one of them The girls one is a bow something But pray I'm gonna need that second Boba bottoms. No Boba tons. Yeah, there's the same thing. You said bottoms Boba tons. Okay, Boba tons And what's the other one? It's it's it's just one. That's what I said one. I'd like to know if you need a little Durham slang or something terms trying. Yeah name name the deathly Hallows Wow Resurrection stone. Yeah, the invisibility coke. Yeah, and I only want No Okay, what position did James Potter play when he was on the Gryffindor quidditch team? I think he's a secret here with me. Yeah What are the letters that are delivered in red envelopes that yell at their recipients called? Uh Howlers, you saw a few howlers last night. Yeah There's a secret passage under Hogwarts. Where does it start and where does it end? and It starts at Big tree thing. Oh, I don't need a name of the tree Eugene. What's the name of the tree? I like that answer the big M What is the tree called?  Sorry 2d. He said that already shrieking truck last question machine Where is the opening to the Chamber of Secrets? It's in the toilet Wait, where the girls? Is the girls toilet which means sign up my friend?  Yeah, you definitely scored more than that like I don't know. Yeah, sir. I'm kind of a mistake of that Yeah, so we don't have an official score Yeah, 19 you got Put it right there. Thanks, man. And listen Ronan.


Wow, yeah, I swear to God I'm with it. I don't see nobody in my lane It's quite go get it like me Please don't me waste of my time with that business to episode number two of can I drop 20 pounds in six days just like The BMF all hay man's video now I'm in the changing rooms. You might have seen in the first clip. We're outside of snap fitness here in New Briden This is going to be my local gym. This is going to be my local Scales and this is going to be where we weigh in on a daily basis for the next six days The time is 7.52 a.m. On November the 4th. I weigh myself the first time in nearly two weeks I competed in a bodybuilding show Since then I have binged every single day on whatever I want. It's not so much protein So I find myself here and I'm going to weigh myself for the first time in a week and a half This is going to be day number one. This is going to be my first I guess check-in with you guys and from here. I'm going to try and drop 20 pounds in six days It's possible.  So my body needs this and I'm going to use this as an opportunity to make another YouTube series So I hope you guys enjoy.  I watch most UFC cards even fight nights and You know, I've got a huge interest in it So when I when I saw the all-hat Nazgadal dropped 20 pounds in six days I knew in my head with my experience having cut a lot of times before I knew that depending on what position his body's in That's definitely possible You know, he's got he's probably got at least five kgs of fat. He could lose in that time Well, maybe about four and the rest would be water manipulation. So that's how I'm going to go for it That's how I'm going to do it.  I'm excited. I really am.  My body feels heavy. My legs feel heavy right now.  I'm going to feel a whole lot lighter. I'm going to feel like a brand new person That's the plan. So the first thing we need to do. I'm fully clothed We're going to take a way in with my clothes on. I'm going to take my clothes off I'm going to take another way in And then we're going to get into the first workout of the week So please go I've stolen the scales. This is me This is what? I'm looking rather large Feeling rather large, but let's get the first way in man Like I said, I weighed 88 kilograms a week and a half ago. So let's see what it's what it's at I'm probably going to cry. I may cry. What does this say? Oh for fuck's sake Zubin 97.3 kgs.  It's 95. Let's see. Look to see. This is me Where did my abs go? They're still there actually. You see this is the point man if I had pushed this bench for any longer This would have completely and utterly blown out and I just cannot let that happen So that's why this video series being made. Okay, no clothes This is how the body's looking and I'm going to take our first way in of the series Now excuse the feet Okay, that's surprising 96.4 kilograms. All right now we're going to log in With my Fitbit, which I will be using the entire time And by my watch that's 758 November the 4th 2020 750 758 November the 4th 2020 What does that say November the 10th? We'll call it. We'll call it 8 o'clock November the 10th. I have to wait What was that 96.4? It's 10 pounds 20 pounds is about 9 kg So I'm thinking I'm gonna have to be down below the weight that I was a week and a half ago How that's gonna happen a bit of discipline, a bit of hard work and a bit of sweat And we're gonna start the sweat right now for our first workout of the fucking week. I Just want to sweat.  I feel heavy. I feel sluggish. I feel bloated Let's start this thing off right and Sweat out a good a 2 kilo Let's see what happens. I'm gonna go and work out. I'm gonna meet you back here And I'm gonna get away and post-work out and that's probably gonna finish us off for the second episode of the series Thanks for being here. Session number one in the books I don't know if you can Don't know if you can see but I'm drenched and that's exactly what I wanted. I'll tell you what Walking out of the gym and feeling the the fresh Cool air on my sweaty face was actually an amazing feeling guys, and it's one that I've missed for the last week and a half so Funnily enough, you know Fuck I didn't I didn't even know if this was gonna happen I did not know if I could ever stop eating and in fact, I don't know if I'm gonna be able to stop because I'm gonna go home now and You know the cupboards are still stocked So obviously it's gonna be disciplined but what I do want to say and what I want to mention before going is that I'm actually using this YouTube channel and this you know this YouTube video right now as Help, you know, it's it's it's support for myself to to begin a Such a thing like this, you know if it wasn't for the YouTube channel if it wasn't for the ability to Share a message to create content.  That is just about Using it as motivation. I guess you could say so, you know anything I do in my life. That's That I'm probably struggling for motivation to start Beginning a YouTube at beginning beginning a YouTube series like this helps me with that and you know And I'd assume it would help you guys too So if you ever thinking about doing something if you haven't started a YouTube channel, you know begin your YouTube channel with a Video series tracking and documenting something that you're doing.  Peace out everybody. Have a great day


So, good morning, students. So we were doing the exercise five of circles in which I think I have completed up to a question number eight, right? So, we will start with question number nine today and we will finish this exercise today. Right. So, let's see question number nine. It is saying that the length of tangent from. It means zero comma zero to the circle. This is okay. So, equation of circle is given here as two X square. Two into X square plus Y square plus X minus Y plus five equal to zero right and it is asking for the length of tangent from zero comma zero. So, we know the length of tangent, the length of tangent is given by under root S one where S one is the power of the given point, right? But for this formula, we need to make the coefficient of X square and Y square as one right then only this formula is applicable. So, what I will do, I will make the coefficient of X square and Y square equal to one for that I have to divide this whole equation by two. So, dividing that, what we get X square plus Y square plus one by two X minus one by two Y plus five by two is equal to zero, right? Now, we can apply formula. Now, we can apply this formula, what we know that is the length of tangent is equal to under root of S. So, put the value of this, what you say this coordinate here. So, putting zero comma zero here, zero comma zero in this equation of the circle. We get this will be zero zero zero zero plus five by two. This will be the length of tangent, right? Length of tangent from zero comma zero. That two, that will be under root, right? So, it will be coming up to be under root of five by two, okay? So, this option B is correct. Option B is correct. Now, let's take the next question, question number 10. It is saying the perpendicular tangents to the circle X square plus Y square is equal to A square meet at P, okay? Then the locus of P has the equation. Okay, means the angle between the tangents are 90 degree, right? Angle between the tangents are 90 degree. So, actually this will, the locus of P will resemble the director circle, right? Because the tangents are meeting at 90 degree, okay? Let me first draw the sketch for this. So, like, this is our point P, okay? This is our point P, where this and the angle between them are 90 degree. So, if you remember, when the tangents meet at 90 degree, so this locus of P, suppose I'm taking its coordinate as h comma k, the locus of P is basically the director circle. It's basically this locus of P will be director circle. And what is the director circle? Means what is the equation of the director circle? Basically, director circle is co-centric, like on, it has same center as the given circle. What is the center of this circle, given circle? This is 0 comma z. So, the center of director circle will be same as the center of the given circle. But the radius of director circle is basically root 2 times the radius of the given circle. So, we can simply write the equation of director circle as x square plus y square is equal to, what is radius of given circle here? A, so it will be root 2A. This will be the radius of the director circle. So, from here, we get x square plus y square is equal to 2A square. This will be the locus of P. So, if you have an idea of the director circle, you can simply write the locus of P. But let's see like how this thing arrives, like how this locus of P arrives. So, it's simple basically. If you see, I'm taking the coordinates of P as h comma k, right? So, what I will do, this is our center of the circle, right? This is our, this is center of circle, which is nothing but 0 comma 0, okay? What I will do? I will join this P to C. I will join these two points, P and C, okay? And I will drop perpendicular from C on these tangents also, okay? So, if you notice this line P, C, this line joining P and C, you actually bisapes this angle. It actually bisapes this angle means what? This angle will be 45 degree, right? And this angle will also be 45 degree. So, if you see, I'm naming it as A and B. So, if you see in triangle P, A, C, if you see in triangle P, A, C, okay? 1045 degree will be equal to AC upon, AC upon, this will be 90 degree nu. So, 1045 will be AC upon PA, right? And what is AC? AC is nothing but the radius of the circle, which is A here in this case. So, AC, we can write AC as A and what will be PA? PA is the length of tangent, right? This is the length of tangent from P. So, this will be under root of S1, okay? So, we can say this under root of S1 is equals to 1045 is 1. So, under root of S1 is equal to A. And what will be under root of this S1?  This whole thing will be under root and this will be equal to k. So, squaring both sides, what we get?  Now, replace this h and k by x and y. So, finally, we get this x square plus y square is equal to 2 square, right? This will be the locus of given point P, which I told you earlier only. This is the equation of director circle. This is what we used to call this locus of P. We used to call this as director circle because y, because the equation is coming. As the circle only and the radius of that circle will be root 2 times the radius of the given set. So, this will be our answer. So, in option if you see, this option A is correct. Option A is correct, okay? Now, let's move to the next question. Question number 11, okay? It is saying the tangents to x square plus y square is equal to A square and inclinations alpha and beta intersect at P, okay? Equation of circle is given and it is given that the pair of tangents intersect at P. If cot alpha plus cot beta is equal to 0, then the locus of P is, okay? So, same thing we will draw once again. So, this is our circle. This is our circle. Okay, this is our point P. And we have to find the locus of point P, right? So, this is our point P. I am taking its coordinates as h from a k. This is our circle x square plus y square is equal to A square. And this inclinations, suppose the inclination of this line is alpha. This is basically with respect to positive x axis, right? So, we can say that it's a slope. The slope of this line is basically tan alpha, right? Since inclination is given as alpha. And its inclination, the line of the inclination of this line is given as beta. So, it's the slope will be basically what tan beta, okay? So, if you see, if I have to write the equation of tangent from P, okay? On this circle, how can I write it? I can write it as y is equal to mx plus minus c. C is nothing but A into 1 plus m square, right? This is the condition for tangency, no? This is the condition for tangency where c square must be equal to A square into 1 plus m square. This is the condition for tangency, okay? So, I am writing the equation of tangent as y is equal to mx plus minus A into 1 plus m square, right? Because the two values of c will come from here, plus minus A under root 1 plus m square. Now, this line is passing through h comma k, right? So, I can replace this x and y by h comma k. So, we can write the equation, rewrite the equation as k is equal to mh, right? Plus minus under root A, sorry, plus minus A under root 1 plus m square. This is the equation of tangent, right? Equation of tangent, which is passing through P, okay? Now, let's make it a quadratic in m. So, what I get? I will be having k minus mh whole square is equal to A square into 1 plus m square, right? So, open it, it will be k square plus m square h square minus 2 khm is equal to A square plus A square m square, okay? So, now, let's take the coefficient of m square together. We get this h square minus A square into m square, okay? Then, I will get minus 2 khm into m, okay? And then, plus k is square minus A square, okay? So, we got quadratic in m. What we got? We got quadratic in m. So, obviously, from here, we will get the two values of m, means we will get two values of m, right? From here, if you see, our m1 plus m2, our sum of roots for this quadratic will be minus B upon m.  And what is m1 and m2? We are given the slope of tangents, right? Here, the slope of tangents, if you observe, we are provided with the slope of tangents in the question itself. So, we can replace this, we can write this m1 plus m2 as 10 alpha plus 10 beta, right? So, the value of 10 alpha plus 10 beta is coming out to be this 2 k h upon h square minus A square. Now, one more information is given here, this cot alpha plus part beta, okay? So, let's make it in terms of cot alpha and cot beta. I will get 1 by cot alpha plus 1 by cot beta is equals to same this thing, 2 k h upon h square minus A square. So, taking LCM, what we get? cot alpha plus cot beta upon cot alpha into cot beta is equals to 2 k h upon h square minus A square. Now, this thing is given as 0, right? This is given as 0 in the question itself. So, from here, if you see what we get, we get 2 k h is equals to 0, right? Now, for locus, we know that we replace this h and k by x and y. So, replacing this, what we get? 2 x y is equals to 0 or we can say x y is equals to 0. So, this will be the locus of point p. This will be the locus of point p. So, x y equal to 0. This way option C will be correct. Hope this is clear to all. This is a good question, basically. So, let's take the next question, question number 12. It is asking the exhaustive range of values of A such that the angle between the pair of tangents drawn from A comma A to the circle lies in the range pi by its way to pi. Okay, means one circle is given here. And we have to draw a pair of tangents, but the angle between the pair of tangents is limited to pi by 3 to pi. Then we have to find the value of A. Obviously, it will lie within some range. So, let's see how can we do it. So, I'm drawing this purple. Okay, and from point A, we have to draw a pair of tangents, right? From point A, we have to draw a pair of tangents. This angle is given. So, I'm doing one thing. I'm joining the center also because later on we have to do this construction and I'm joining these points. So, this is all. Let me name it as point P. Okay, whose coordinates are given as A comma A. Let me name this point of contact as AB, this center as C. Okay. So, given, if you see the equation circle, what will be the coordinates of this C? It will be 1 comma 1, right? It will be 1 comma 1. And what will be the radius of the circle? It will be 1 square plus 1 square minus C minus A is means minus of minus X that will be plus 6. So, radius will be 1 plus 1 to under root 8. The radius of this circle is under root 8 and center is 1 comma 1. Now, it is saying that the angle between the pair of tangents should lie in the range pi by 3 to pi. Okay. So, suppose I'm taking this angle as 2 alpha, right? So, this 2 alpha as per given condition, this 2 alpha should lie between pi to pi by 3. Okay. If this angle, this angle between pair of tangents is 2 alpha, what will be this angle? This angle will be alpha, right? Because this PC will bisect this angle. So, this will be alpha, this will be alpha. Now, we will play with this alpha means, so, if 2 alpha is lying between pi by 3 to pi, where should this alpha lie? Alpha should lie between pi by 6 to pi by 2, right? And if I say, where should sine alpha lie? Sine alpha should lie between, what is sine pi by 2? It is 1 and what is sine pi by 6 means what is sine 6, sorry, sine 30 is 1 by 2. Okay. So, this will be the range of sine alpha as per given condition. Sine alpha should lie between 1 by 2 to 1. Now, what is sine alpha? From this figure if you see, what is sine alpha? Sine alpha will be, this will be a right angle triangle, right? So, sine alpha will be basically AC, right? AC upon PC. Okay. And what is AC? AC is nothing but the radius of the circle. So, I am representing at R and what is PC? We can find this length of PC since coordinates of A and C are known to us. So, if you see, I am writing here, PC will be nothing but under root of A minus 1 whole square, right? Plus A minus 1 whole square. So, this is nothing but 2 times, 2 times A minus 1 whole square under root. Okay. So, this will be under root of 2A minus 1 whole square. Okay. Now, sine alpha is lying between 1 by 2 to 1, right? So, I can say this thing, put the value of sine alpha here. So, let me write in terms of AC and PC only because it will be short. So, R upon PC should lie between 1 by 2 to 1, right? From here, if you see, this is an inequality where we can see, multiply it by 2. It will be 1, 2 R upon PC should be 2. So, from here, we get two conditions like this PC should be greater than R and PC should be greater than R and 2R is greater than PC, right? Or PC is less than 2R. So, PC is greater than R, but it should be less than 2R. Okay. Now, what is PC? PC, we know the value of PC, length of PC. So, it will be under root of 2A minus 1 square should be greater than R. Okay. So, square it, we will have two times A, A square minus 2A plus 1 should be greater than R. R squared will be 8. Okay. So, this will be cancelled out.  So, factorize it. It will be A square minus 3A plus A minus 3 greater than 0. So, A, A minus 3 plus 1, A minus 3 should be greater than 0. Finally, we get it as A plus 1 into A minus 3 should be greater than 0. Okay. Now, find the solution for this inequality. So, let me draw the critical points. It will be minus 1 and plus 3, right? So, anything value of A greater than 3 will give me positive result. This will be negative and this will be negative, but what we need, we need positive. So, we get the value of A as minus infinity to minus 1 and union 3 to infinity. But our work is not yet done. We have to solve this inequality also. PC is less than 2R. So, solving this, if you see, to under root of A minus 1 whole square, right? This is PC. This should be less than 2 times R, that is 2 under root 8. So, cancel it. Oh, no. This will be under root, no? Oh, I have written it wrongly. So, this will be, let me erase it first. So, what I have to do? This PC is nothing but under root of 2A minus 1 whole square, right? This is PC. This should be less than 2 times under root 8. Now, square it. So, we get 2 into A square plus 1 minus 2A should be less than 4 into 8, that is 32, okay? So, it will go 16 times. So, finally, we get quadratic as A square minus 2A plus 1 minus 16, that is minus 15. This is less than 0. So, split it, A square minus 5A plus 3 minus 15 should be less than 0. So, A into A minus 5 plus 3A minus 5 should be less than 0. So, finally, we get A minus 5 into A plus 3 should be less than 0, okay? Plot the, now plot that critical point here for this equation, okay? What we get? This will be minus 3. This will be our plus 5. So, anything ever 5 will give me positive result. So, this will be positive. Anything between minus 3 to minus 5 will give me negative and anything less than minus 3 will give me positive result. But we need negative. So, from here, we get A belongs to minus 3 to 5, right? Now, we have to take the intersection of both these solutions, right? We have to take the intersection. So, finally, we get AS, finally, we get AS, A belongs to from minus infinity to minus 1. So, minus 3 to minus 1, right? And union, here it is going from 3 to infinity, but limiting point is 5 only. So, it will be up to 5 from 3 to 5. So, this will be our final range of A for which conditions such that the angle between the pair of tangents will be in the range of pi by 3 to pi. So, this will be our answer. Option, option, if we take the option, option B is the left, okay? So, this was our question number 12. Now, let's move to the next one. Okay. So, it is seeing distances from the origin to the centers of the three circles, x square plus y square minus 2 lambda x equal to c square, where c is constant and lambda is available. No, it should, it should be variable basically. So, let me correct it. So, it is saying that lambda is variable, variable. Okay. So, a lambda is variable, R in GP means distances from origin to the centers of circle are in GP, okay? So, I think that the lengths of tangents drawn from any point on the circle, x square plus y square is equal to c square to the three circles are also in GP. Okay. So, I think question is clear to everyone. Like distances from origin to the centers of the circles. Let me write the equation of the circles. So, write the equation of first circle. So, it will be like x square plus y square minus 2 lambda 1x since lambda is variable. So, for circle 1, I'm taking lambda is lambda minus x minus c square is equal to 0. Similarly, we can write the equation of circle 2 and x square plus y square minus 2 lambda 2x, okay? So, minus c square is equal to 0. Rest all things with c will be constant since it is constant. And we can write the equation of circle 3x, x square plus y square minus 2 times lambda 3x minus c square is equal to 0. Okay. So, tell me what will be the center for this circle. If you see what will be the center for this circle, it will be basically lambda 1, 0. What will be the center for this circle? Okay. What will be center for this circle? It will be lambda 2, 0. And circle for this center for this third circle will be lambda 3, right? Now, distances of these points, distances of these points from center is constant, not constant, RNG. Okay. Similarly, let me assume this origin as O. So, this O c1. O c1 will be lambda 1 minus 3 to the square means distance formula. So, it will be under root of lambda 1 square that is nothing but lambda. Similarly, our distance of second center means the center of second circle to origin will be under root of lambda 2 square that is nothing but lambda 2. So, distance of this center of third circle to origin will be lambda 3 square that is lambda 3. So, these are the distances as per question it is given that this lambda 1, lambda 2 and lambda 3 is in GP, right? Now, what we need to prove? We need to prove that the lengths of tangents drawn from any point on the circle. This given circle x square plus y square is equal to c square to the three circles are also in G. Okay. So, one circle is given here, x square plus y square, x square plus y square is equal to c square. Let me assume one point on this circle as, let me assume one point on this circle as p whose coordinates we can give it as c cos theta. r cos theta comma r sin theta, we used to take any point on the circle in parametric form in this way only. So, I am taking one point on this circle as c cos theta comma c sin theta, right? Now, I will take, means we have to draw the tangents from this point on the three circles, okay? We have to draw the tangents from this point to the three circles, okay? So, suppose I am taking it as, now I am calculating the length of tangents, okay? So, this length of tangents, I will calculate length of tangents from this point on the circles, okay? So, how can we do it? For first circles, let me denote it as under root of S1, okay? So, what will be our under root of S1? This x and y wherever this x and y are there in the equation of these circles, we will replace it that x and y by this c cos theta and c sin theta. So, it will become c cos theta whole square, right? x square plus y square, y we will replace it as c sin theta square, right? Minus 2 lambda 1 into c, sorry x, x is nothing but c cos theta, right? And minus of c square, this whole thing will be under root. So, if you see here, this will become c square cos square theta plus c square sin square theta. That will be nothing but c square, then minus 2 lambda 1 c cos theta, okay, minus c square. So, this c square, this minus c square plus c square will be cancelled out. And look at this, here I can assume this thing, this minus 2 c cos theta, okay? I am writing it separately and multiplied by lambda 1, okay? Now, observe that this will be some constant, right? What I have written in this bracket, it will be some constant, some constant k. Why? Because c is also constant, 2 is constant and this cos theta will also be constant because what I have done, I have taken one point on this circle, okay? Whose inclination with the x axis is theta? That is constant for this point. So, we can say it as some constant as k. So, this under root a1 is coming out to be under root of, under root of some constant, we can place it outside also. So, it will be nothing but this k into under root of lambda 1. So, this will be our length of tangent on the first, right? And similarly, the length of tangent on the second circle will be k into under root of lambda 2. And our length of tangent on the third circle will be k times under root of lambda 3, right? And we know that this lambda 1, lambda 2 and lambda 3 are in GP, right? It is given in the question. So, obviously, if lambda 1, lambda 2 and lambda 3 are in GP, right? So, this thing, this S1 means k into this lambda 1 or k into under root lambda 2 and k into under root of lambda 3. This will also be in GP, okay? You can take any values and you can also, this if lambda 1, lambda 2 and lambda 3 are in GP. Their roots will also be in the GP. So, we have proposed this. Hence, this is what we were asked in the question. Okay. So, now we will take this next question, question number 40. It is saying that they find the area of the quadrilateral formed by a pair of tensions from a point, from this given point, 4 comma high to this given circle and the pair of its radii. Okay. Same sketch we will draw once again. These two centers. And we will join this point. We will join this point to the center. Okay. So, let me name it as P, this point of contact says AB and this is the C center of the circle. Okay. Now the coordinates of this point P is given as 4 comma 5. And we know the equation of circle as X square plus Y square minus 4 times minus 2 Y minus 11 is equals to C. The center of the circle, it will be 2 comma 1. Right. So, we know the coordinates of this C as 2 comma 1. Okay. And what will be radius for this circle? It will be 2 square means 4 plus 1 minus C minus of minus 11. That is nothing but plus 11. So, it will be 4 plus 1 pi 11 plus 5. So, 11 plus 5, how much? Have I done some mistake or what? So, center of this circle will be 2 comma 1. 2 comma 1. Right. And it will be 12, 16. Okay. Radius is coming out to be 4. Okay. Route 16 is nothing but 4. So, radius is 4. So, we got this radius as 4. AC is out, nothing but radius only. So, the length of AC is 4. Okay. This will be a right angle triangle. So, question is asking to find the area of this quadrilateral PACB. Okay. But we already know that this line PC. Both this, if I write it as triangle PAC and triangle PBC, both will be congruent. Why? Because these angles are equal. These sides are equal. And this is the common side. So, side angle, side congruence, we can show here. So, anyhow, if we find the area of this triangle PAC, area of triangle PAC, and we will double it, then we will find the quadrilateral serial. So, area of triangle PAC, if you see, area of triangle PAC is equal to half in base, like this is a, what do you say PA, right? Half base into height, that is AC. Okay. And how can we find this PA? This is the length of tangent basically. So, from here, we can find PA. PA is under root of S1. We know the coordinates of PA. So, I will put there. So, this will be 4 is where that is 16 plus y is where that is 25, minus 4 into 4 minus 16, minus 2 into y, that is minus 10 and minus 11. So, how much it is coming? 16 and 16 will be cancelled out. This 25 minus 21, that will be 4, means 2. So, PA is coming out to be 2. Now, put the value of PA here. So, it will be half into AC. AC is our 4. So, this is 4. So, area of triangle PAC is coming out to be 4 square in it, right? So, our area of quadrilateral will be, area of quadrilateral, PACB. PACB will be 2 times 4, that is 8 square in it. Okay? So, we got the area of quadrilateral as 8 square in it. So, this was our question number 14. Now, let's take this question number 15. It is saying if the length of tangents from a given point, whose coordinates are f from a g to a circle this, before tense the length of the tangent from the same point to the circle this, x square plus y square is equal to 4x. Now, the question is asking to show that this, this, this is equal to 0. Okay? We know the point that this point P whose coordinates are f, we have to find the length of tangents from this point on these two circles. Okay? So, we know that length of tangent is under root of S1, that is under root of the power of the given point. So, it is given as under root of S1 is 4 times under root of S2, right? To the circle, we 4 times the length of the tangent on the another circle. So, what will be S1? If you see here, S1 will be basically, I will replace this x and y by f and g. So, S1 will be f square plus g square minus of 4. This will be our S1, okay? And what will be our S2? S2 will be f square plus g square minus 4 into minus 4 into x, x is f. So, this is what S1 and S2. Now, put the value there and square it. So, we get this S1 square is equal to 16 into 16 times S2, right? So, S1 square will be, sorry, this square will not come. This is already in the under root, no? So, S1 will be basically equal to 16 times S2, okay? So, put the value of S1 now. It will be f square plus g square minus 4 is equal to 16 f square. f square plus 16 g square minus 16 into 4, 64 f, right? Now, take all the terms to the one side. So, this 15, 16 f square minus f square, that will be 15 f square. 16 g square minus g square plus 16, 15 g square, right? And this minus 64 g plus 4 will be equal to 16. This was, this is what asked in the question. So, we have proved that, yeah, the value of 15 f square plus 15 g square minus 64 f plus 4 is coming out to be 0, okay? So, let's take this question. Find the equation of that part of the circle, which is bisected at 3 comma 2. So, one circle is there and its chord is given as, suppose one chord is given here, which is bisected at 3 comma 2, right? So, it's middle point. So, suppose this is AB. So, it's middle point, okay? This middle point, the coordinates of this is given as 3 comma 2. And now, we have to write the equation of this AB. We have to write the equation of AB, okay? Whose middle point is going to us, which is 3 comma 2. So, if the middle point of a chord is going, so we basically write the equation of AB as T is equal to S1. This is what the formula says, T is equal to S1. And what is the equation of circle? This is x square plus y square minus 15 is equals to 0, okay? So, what is T? Basically, normally, we write the equation of tangent putting the coordinates of this midpoint, okay? And what is S1? S1 is nothing but the coordinates of this point, it means power of this middle point. So, in this way, we find the equation of this AB. So, what is T basically? How do we write the equation of tangent? We replace x square by xx1, right? So, this will be xx1 and replace y square by yy1. So, yy1 minus 15 is equal to S1. Now, what is the power of point? We put the coordinates in the equation itself. So, x square means 3 square, y square means 2 square and minus 15, right? So, this is what we got. Now, x1 is equal to 3 for this and y1 is equal to 2. So, take it. So, this will be 3x, okay? 3x plus 2y minus 15 is equal to 9 plus 4 minus 15, okay? So, this will be 3x plus 2y minus 15. And what is this? 9 plus 4, 13 minus 15, that is minus 2. It will become plus 2 when it comes to this side. So, finally, we get the equation of abs, 3x plus 2y, sorry, this 3x, 3x plus 2y minus 13 is equals to 0. So, this is our equation of ab. This is what we were asked to find, okay? Now, let's take the, I think one more question is left in this exercise. Yeah, this question number 17. So, it is saying the quads of contact of the pair of tangents to the circle x square plus y square is equal to 1, drawn from any point on the line this, pass through the points alpha, comma, beta, then alpha square plus beta is broken. So, one similar question we have done in this exercise itself. This includes the concept of what you say, family of lines, right? So, this question is also based on the same concept. So, this is our pair of tangents, which we are drawing here, okay? And we have this point is lying on the given line, what is given in the question. This is lying on one line this, whose equation is known, it was, okay? So, let me take this point as p, okay? As usual, we are taking this point as p only and on the line, this equation of this line is 2x plus y is equal to 4, okay? So, this point p is lying on this line. Like this point is variable, this point p is variable, it can lie anywhere on this line, okay? So, after, like, how can we write this coordinates of p? If we say, we can rewrite it as this equation as y is equal to, right? y is equal to minus 2x plus 4, okay? You can write this equation, 2x plus y equal to 4 is given. So, we are writing this as y is equal to minus 2x plus 4. So, suppose if I am taking this point as h, the x coordinate as h. So, what will be the y coordinate? Y coordinate will be basically minus 2h plus 4. Is it clear? Because this point is p is lying on this line only. So, anywhere it lies on this line, the coordinates of this point p will be of this formula, h comma minus 2h plus 4. Okay? Now, question is saying this point of contact. Let me join this also. Okay? So, it is saying the codes of contact. Look here, it is saying the codes of contact. It will be not a single line, it will be a group of lines or a family of lines. So, the codes of contact, this is our code of contact, okay? Of the pair of engines to the circle. This is circle x square plus y square is equal to 1 drawn from any point on the line. Okay, we have taken that point, passes through point alpha comma beta. So, this code of contact should definitely pass some fixed point alpha comma beta. And we need to find that point. We need to find that point alpha comma beta. Then only we can find this alpha square plus beta square. So, let me write the equation of AB. How can we write the equation of AB? How can we write the equation of AB? P is equals to 0, right? Where the tangents, like the equation of tangents to be drawn assuming the point as this, coordinates of P. Okay? So, our equation is given as xx1 plus yy1 plus yy1 minus 1 equal to 0, okay? So, x into x1, what is our x1? That is h and y into y1, what is our y1? 4 minus 2h, right? 4 minus 2h minus 1 equal to 0.  Okay? So, we can rewrite it as 4y minus 1. I am taking these two terms together. So, this will be 4y minus 1 plus h into x minus 2y, right? So, what does this represent? This is representing the family of these lines, means family of this chord of context, which are passing through the intersection of these two lines. Suppose I am taking this as l1 and this as l2. So, this family of lines will definitely pass through the intersection point of these two lines. Okay? So, now we have to solve this. This 4y minus 1 equal to 0. From here, we get y equal to 1 by 4. And if we get y equal to 1 by 4, put it here. So, this x minus 2y equal to 0. Now, x is equals to 2 times y. That is 2 into 1 by 4. That is nothing but x is equals to 1 by 2. So, we got the point. Means the chord of context will definitely pass through this point c, whose coordinates are 1 by 2 comma 1 by 4. So, we got the value of alpha as 1 by 2. And the value of beta as 1 by 4. Now it is asked to find the value of alpha square plus beta square. So, what will be our alpha square plus beta square? It will be 1 by 4. 1 by 4 plus 1 by 16. That is nothing but 4 plus 1. That is 5 by 16. Right? This will be our value of alpha square plus beta square. Okay. So, hope this exercise is clear to all of you. So, now this, we are done with this exercise. We will be back one soon with the next exercise that is exercise number 6. Okay. So, till then, take care.