there becomes difficult. Anyways, so let's begin with circles. And I have again, you know, brought to you this slide which talks about the mark distribution. Okay, so you can see we have completed algebra 20 marks. We have completed trigonometry 12 marks. So 32 in all. So you can see number systems is six for R and R people. There is no number system. Instead, they have rational expressions and matrices. So that's different for them. Coordinate geometry, six marks, geometry, 15 marks. This is where we are going to begin today, geometry. And then it's area volume and all that. So that's when menstruation, 10 statistics and probability 11. So this is how, you know, so you can see in geometry, again, you have which all chapters. You have triangles, the most important one or the most dreaded one as well. Most of the people actually do not feel themselves comfortable find themselves comfortable in triangle. So we'll spend two classes there. We'll go slow. We'll talk at a very detailed manner triangles. And because there is case study on triangles, most likely there will be a case study on triangle chapter as well. So hence that is important chapter. Then construction and then this circles. So we are going to take up circles today. Okay. So again, there is no deletion. Just for your information, there has been nothing which has been taken away from the circles chapter. So you have to go through the chapter in full. So here there is no concession made to you. Okay. Now again, the ritual is to go through the, you know, the structure of the exam. There are two parts, AB 36 plus, sorry, 32 plus 48 is the breakup, 36 is 16 plus 16, 16 one marker. And a four case studies with five questions each, but you have to answer only four of them. So 16 plus 16, 32. And then there are six two markers, seven three markers and three five markers. So this is the breakup. And internal choices is provided in two questions of two marks, two questions of three marks and one question of five marks. So you'll get three choices in the question pick. Okay. And the four choices in the case study. So please remember that. Okay. Okay. So let's begin. So here is the definitions.  And circle is a locus. Now people do get confused when I talk about things like locus. So, so this is locus. What is a locus? So circle is a locus of a point in a plane, which is at a constant distance from fixed point on the same plane. So here I have tried to show you a small, this is what it means. So there's a point which is going around a point such that the distance between that moving point and the, so once again, the distance between point P and A is always constant. And if you, if the point P is tracing that path, that path is called the circle, right? So that's locus. The word is locus Greek origin. Locus means path, path of a point in a plane. So that's important because you can have similar such figures in 3D also. So if you do not restrict it to a plane, it will become a sphere. So a plane which is at a constant distance from a fixed point on the same plane. Okay, this is important. The constant distance is called radius. You have studied since what? LKG, UPG kind of thing, no. Or in which grade did you study radius first? So very early stage primary schooling and center of the circles. So you know this, so no spending time here. Sorry, next, yes. So now in the ninth grade, you studied more in detail the properties of circle, like chords, like, you know, what else? The arcs and the theorems related to that, right? The perpendicular to the chord and all that. In 10th grade, we took up, took another two vital component of the circle, which is, or a part of circle you can say, secant and tangent. What is a secant? So secant is nothing but a line which intersects a circle in two distinct points. So if it is intersecting in two distinct points, then we call that secant. So you can see PQ here in the diagram here, P and Q. I'm not able to hear you. Who's this? Sir, it's Aditya. Yeah Aditya, what happened? You're not able to hear me? Hello? Am I audible? Do you now? Yes, everyone? Audible? Am I audible? Yes. Okay, cool. So wait. Okay fellows. So what I was saying is, secant definition is clear to you. So every, how many secants are possible then for a circle? How many secants can you draw to a circle? How many? How many circles can be drawn to, oh sorry, how many secants can be drawn to a circle? Okay, if I restrict how many to this question, how many secants can be drawn from one point on the plane? One point on the plane to the circle. How many secants? So let's say if I define a point here, this point. Oh, I don't, let's see this point. This is visible. P or Q. So how many secants can I draw onto this circle through this infinite secants? Very good. What if the point is on the circle? How many secants can be drawn? Infinite again. And what if the point is within the circle? How many secants can be drawn? So in all the cases, there are infinite amount of secants you can draw onto a circle. There is no restriction. There are infinite amount of parallels, parallel secants as well, infinite amount of mutually perpendicular secants and things like that. So secants are infinite. Okay, very good. So what's the tangent? Tangent is a special case of secant. So what's that special case? So wait, if I have to also show you. Through this, yeah. So let's say we have a circle with any given radius. So let's say this is the center and we take a radius of five. Okay, what happened? I don't know. So wait, okay. So we need this. So what I'm saying is I'm drawing a circle. Circle from this point to let's say this point. And then I am drawing a secant. Secant is nothing but to take any two points. This is a secant. Now if I move this point and let's say, let these are the points of intersection E and F. And what I'm going to do is I'm going to move this D point. So as I'm moving this point D, so you can see there, these are all secants, but now E and F are moving together and there will be a point when they will get merged together and that is where it becomes the tangent. Oh, this is just losing the connect. Yeah, so this is almost, it's very difficult to actually restrict it to that, but let us say, yes. So when E and F are merging together, right now also this is a secant. So there is some gap between the circle and, yeah, so you can see I can move this and move this. So this is secant. Okay, secant and tangent. So tangent is nothing but the limiting case when the two points merge together, they become tangent to the circle, right? So how many tangents now? So this is the question. Tangent can be considered as a special case of a secant when the two points of intersection coincide. Okay, now, so inside the circle, there are zero tangents possible, right? So from inside of a circle, you cannot have any tangent drawn onto the circle. If the point is on the circle, then one tangent is possible. As you can see, if you consider this point E here or F here, only one tangent is possible. And if the point like D is outside, then there are two tangents possible. And the entire 10th grade syllabus around the circle is dependent on the properties of these tangents and secants. And tomorrow when you go to 11th grade, when you talk about, let's say, when you will be doing conic sections, in conic sections, you will see, again, a lot of application of tangents and normals and secants and things like that, okay? So let's go ahead. So what is the length of the tangent? Now we are defining something. It's called length of a tangent. What is length of the tangent? So from wherever you are drawing the tangent and the point has to be outside the circle. And if you are drawing the tangent and this point Q is called the point of contact. Now point of contact to the point from where you have drawn the tangent, that length of the segment will be called length of the tangent, okay? So don't think that the length of the tangent will be infinite. No, so how do we define length of a tangent? From the point where you are drawing or you are drawing the tangent and you are measuring the distance between that point to the point of contact is length of the tangent. So let's begin our journey of theorems. There are 10 theorems related to circles which you must know. And in last year, that is in board exams of 2019 March, there was question based on proving. And exactly this time, theorem one was there as a, I think three or four marker will check this. That question is also there down there in the slides. So this is the first theorem. It says the tangent at any point of a circle is perpendicular to the radius through the point of contact. Now this radius is called the normal, right? So the radius, let me just change this color because it's too, okay. So this is called normal. This one, OP is normal. What is it saying? The tangent at any point of a circle. So you can see PQ is the tangent here and they are saying that it will be perpendicular to the radius through the point of contact. Okay, so OP, you have to prove to prove. OP is perpendicular to OQ, okay. Anyone knows how to prove this? That OP is perpendicular to OQ. How do you prove that? So here we are going to use what you learned in ninth grade, Euclid's some theorems and some, you know, postulates which Euclid had given. And what is that which you're going to use here to prove this particular theorem? You remember, guys, how do you prove that OP is perpendicular to OQ? Anyone can unmute and say, no problem. Yep, anyone? No, no one is willing to prove this. Shortest distance from O is perpendicular on the line. So how do we prove that this is of shortest distance? Stress. So how do you prove that? Yes, so this is the theorem which, or this is the postulate which stress is talking about. So we are going to use that. So anyways, so what we are doing is, we are taking a point on the tangent. Let's say that point is Q. So here is the proof. What is the proof, right? Yes, so let's take, let us consider. So in geometry questions, the majority of the marks are lost in missing steps. So you have to be very, very thorough and methodical while writing geometry answers, okay? So let us consider a point Q on the tangent, okay? So clearly Q is outside the tangent. So clearly, clearly Q is outside the circle, right? Outside the circle, right? Because the tangent by definition touches exactly at one point. So the P is that point. Then obviously Q has to be outside the circle, right? Let OQ intersect the circle, the circle at Q dash. Okay, then clearly OQ is greater than OQ dash. Isn't it? OQ dash, which is equal to radius of the circle R, okay? OQ dash is the radius, OP, right? This is what you can say. So OQ is, and why is this? You have to write the reason sum is, or whole is, whole is greater than, is greater than the part. Okay, so whole is greater than the part. Now, so OQ is greater than OP, right? Right, that means, that means for any other Q. So you can write for any other point Q on the tangent, OQ is always greater than OP, right? So whether you take a point here or here or here, anywhere you take, you will find OQ greater than OP. Hence, hence, OP is the shortest distance. Shortest distance to PQ from O, from O, hence. Hence, what can we say? We can say OP is perpendicular to PQ. Okay, so you have to write a few more lines. You have to mention, if you wish you can mention Euclid's name as well. So this is where, right, any issues in proving this for understanding this proof, right? So I have not written the given and the proof. There was no construction, so hence you just start like that. Below, any difficulty? Guys, any difficulty? Any difficulty? Just give me some, you know, some signal. Yes, yes, no re-expansion needed, no doubt, whatever. Yep, come on guys. Energy, no doubt, no doubt, all clear. So hence, you can keep this in mind whenever this kind of a theorem is there. So you need to, you know, use the shortest distance route to prove perpendicularity. Let's go to the next theorem and it is a converse of the same. And this theorem is used for construction, which anyways we'll see later. What is that? A line drawn through the end of radius and perpendicular to its, it is a tangent to the circle. So what are they saying? A line drawn through the end of a radius. So end of the radius is P and you're drawing up a perpendicular on OP, right? It's already given that it is 90 degrees. So you have to prove, you have to prove that PQ is tangent to the circle, okay? So how do you prove this? So this theorem can be used, as I mentioned, to construct tangents to a circle from points lying on the circle. So later on in the next chapter in the circles, construction chapter, they will give you construct a tangent on a radius, let's say, is a radius and some radius is given. So you have to construct a tangent on it. So how will you construct? So you'll construct the circle, take the radius and then draw the tangent, like that, right? That construction is, anyways, we'll discuss that, okay? So construction, this is what it is used. So who will prove it? Any idea on proving how to prove this one? A line drawn through the end of a radius and perpendicular to it is a tangent to the circle. What up? How will you prove? So you can use the previous theorem. Now for Converse, the best part is you can always use the previous theorem to prove this theorem, is it? So how will you prove? Any idea, guys? Who has done it? How will you prove that? A line drawn through the end of a radius and perpendicular to it is tangent to the circle. So hence, you can use the contradiction root, is it? So what can you say? You can say, let this line, or first write all that basic things. What are the basic things given? What is given? OP is a radius of circle with center O, with center O, okay? And it is given that PQ is perpendicular to OP. To prove, what is to prove? So I'm writing, in short, you don't write it. So to prove, you have to prove that PQ is tangent to the circle, okay? So if a point is there on the circle, so what are we trying to do? We are trying to go through contradiction, you contradict some statement and you prove it. So what are we going to do? So we are saying that let PQ not be the tangent. Okay, fair enough, not be the tangent. So if PQ is not the tangent, tangent, right? If PQ is not the tangent and P is on the circle, that means PQ is a secant, correct or not? If something is not, if something is, if some line is not a tangent and one of the point is lying on the circle, then the line has to be a secant. I agree, all of you, yep. So if PQ is a tangent, then my logic would be then, sorry, is a secant, PQ is a secant. Then what else can be thought of? So if PQ is a secant, then yes, how do we build this proof, follow? Build it, guys, you also participate so that it remains in your memory. What else can be done? So if let's say it's not a tangent or you prove it with the shortest distance again, yet again. Yep, so you use the same funda of shortest distance. What is that shortest distance thing? So either you prove that if this PQ is a secant is also a tangent. So P and Q will merge together, then done, okay? Then you can do that.  You can do that. So if PQ is a secant, then there will be, so this is the logic, yes, that's what, this is what we are going to do here. So what you can do is, let's say if PQ is a secant, then there exists a tangent which will be passing through P anyways, okay? So as you're suggesting, you can call this as instead of PQ, you can P dash, Q dash, coin, PQ dash. So let PQ dash be the tangent. Let PQ dash be the tangent. If PQ is not the tangent, there will be one PQ dash which will be tangent. There has to be one tangent passing through that point. That means what angle O PQ dash has to be 90 degrees. Isn't it? O PQ dash has to be 90 degrees. Why? Because of the previous theorem, what we proved that our tangent and our radius on the point of contact are perpendicular to each other. Are you guys getting this? The build of, the logic, how am I building the logic? I'm saying let PQ is not the tangent. That means PQ is a secant. That means you can pass our tangent from point P to the circle. Let that tangent be PQ dash. So PQ dash is the tangent now. If PQ dash is the tangent, then angle O PQ dash is 90 degrees. But you're also saying O PQ is 90 degrees. Okay, what does it mean? What does it mean? If O PQ dash is 90 and O PQ is 90, that means if Q dash and Q are on the opposite sides, so there are two possibilities.  So two possibilities. What are the possibilities? I will give you two proofs for this. So there are two possibilities. What are two possibilities? So the possibilities are something like this. If Q dash is, so let's say this is O, this is P, either Q dash is on this side opposite to Q. Opposite to Q. And you're saying this is also 90 degree and this is also 90 degree. Then by linear pair, I can say Q dash, P and Q lie on a straight line. Yes or no? Is it it? Yup. And let's say you said no, why are you picking Q dash on this side? Instead of that, you take Q dash here. In that case, if both are 90 degrees, that means Q and Q dash coincide, are coincident. Isn't it? Because if one arm is common of an angle and two other arms are giving you the same angle, that means the two other arms are nothing but they are coincident. That means in the second case, this was one. And in second case, they are coincident. So in both the cases, tangent PQ, sorry, in both the cases, PQ and PQ dash are same lines or coincident lines. Am I right? So now you're saying PQ dash or the tangent and hence PQ has to be a tangent. Is this proof understood? So what did I do? I went for contradiction. So how did I go for contradiction? So I said, okay, let PQ not be the tangent. So if PQ is not the tangent, then that means PQ is a secant. And hence there will be one tangent passing through P. Let that be PQ dash. And then I said, and I used the previous theorem which we proved that OPQ dash will be 90 degrees. And OPQ will be 90 degrees. Both will be 90, this is given. This is given, so I can't deny this. And this is by, this is by construction or let's say the previous theorem. So both are true. If both are true, then that means either Q dash PQ is a straight line or Q dash and Q are the same points. These are the two conclusions. And in both the cases, you prove that PQ and PQ dash are the same lines. And since you assumed PQ dash to be the tangent, hence PQ has to be the tangent. Clear? Proof is clear to everyone. You can also prove it through the larger, smaller, the shortest distance route as well. So you will say that since OP is, so this is another proof, OP is perpendicular to PQ. That means P is the shortest distance. P is the shortest distance. That means any other point Q. So let's say in that case, you will get OQ is greater than OP. Once again, follow the proof. It is given that OP is perpendicular to PQ. Hence, OP is the shortest distance. Then any other point on that line will be greater than OP. So OQ will always be greater than OP. And that is there. That means every other point, so OQ dash plus Q dash Q is greater than OP. That means, what does this mean? This means that Q is an outside point, is a point outside the circle, always. So if all the points except P are outside the circle, because it is immaterial of choice of Q, any Q you take, it will be outside the circle. Then if all the points lying on the line are outside the circle except P, that means there is only one point of contact between the circle and the line. And hence, the line is tangent to the circle. Is the second way of proving clear? So both ways you can prove. Is the second way clear, guys? Anyone without any doubt, please ask, please ask. Okay, let's go to the next one, this one. Now in one itself, we can prove so many theorems related to this. So what does this theorem say? It says the length of tangents drawn from an external point to a circle are equal, okay? So what is meant by the length of the tangent we defined? So length is nothing but they are saying, they are saying this length PQ is going to be equal to PR. So there is a point P and you have drawn two tangents. So how will you prove it? Who will prove this? Anyone? So what is the basis of this proof? That you have to prove that PQ is equal to PR. Yes, so congruence, so basically congruence. So what you'll do is given. And this was also asked in one of the papers. So given is PQ and PR are tangents to the circle with center O, okay? To prove PQ is equal to PR. So what construction? We have to do some construction. Construction is PQ, sorry, OQ, OR and PO joined, okay? Now, so the proof is in triangle, which triangle? PQO, PQO and triangle, PQO, ROP. Have I written correctly? So right now, here it is okay. You can write in this fashion, but is the order of the vertices correct? Is this order, order is correct, PQO and ROP. Are they congruent? Are these two triangle congruent? Yes or no? No, good. So you have to write equivalent. So P or corresponding, PRO, now it is okay. PRO, now it's fine, okay. So PQN, PRO, what do we know? We know that OQ is equal to OR, why? Because these are radii of the same circle of same circle. Very good. What's next? Angle OQP is equal to angle ORP is equal to 90 degree each. And you have to mention that term. So what is that theorem? Theorem is tangent and radius at the point of contact. Contact are perpendicular to each other, okay? So this you have to mention. And what else? OP is equal to OP, common. So by RSA, no, by SSS, true. By SSS is the right, SSS congruence criteria, correct? No, RHS, oh, is it? Okay, good. So by RHS congruence criterion, triangle what? PQO is congruent to triangle PRO. And hence the following, what all? So in one proof, you can prove multiple things. The next theorem also. So you can clearly say that PQ is equal to PR, first of all, done. What else can be said? This can also be said that this angle is equal to this angle, isn't it? All CPCT, so this is alpha and then beta, they're equal. Is it it? Can I say that alpha is equal to beta? Can I say alpha is equal to beta? And what else? I can say this gamma is equal to this delta, right? So alpha is also equal to beta, so this is another proof. And gamma is also equal to delta, these are the proofs. Okay, so the following theorem is regarding that only. See, two tangents are drawn from an external point. This can be given as a question. So they subtend equal angles at the center. So you can see, this is PQ and it is equal, it is subtending angle alpha here. And PR is another tangent and it is subtending an angle, beta here, so alpha is equal to beta, same proof. And similarly, they are equally inclined to the line segment joining the center to the point. What does it mean? So this is the line segment joining the center to the point and both of them are equally inclined. So this is alpha, gamma, and this is delta, right? So we have proved on a previous slide. And PO is the bisector of angle QPR, clearly PO, or O lies two and three mean the same, two and three, two and three, yes. So there are different types of ways they can ask you. So they can ask you prove that center lies on the angle bisector of the angle made by tangents drawn from the same point on the circle, right? So hence, they can give you in different wordings, you have to draw the inference and accordingly solve it. So these are the three cases, which will be there. So one PQ is equal to PR, alpha is equal to beta and gamma is equal to delta. So far so good, any doubt so far? So this is theorem number four, right? So there are questions which are converted into, you know, so theorems which are converted into questions and given to you, next. Now, and there are lots of questions based on this theorem as well. In two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact, okay? So what does this mean? So there are two concentric circles. What are concentric circles? Circles with the same center are called concentric circles in the same plane, that is we are talking on the only 2D case. So there could be circles with the same center but they are not concentric circle. Can you give me an example? Cases where, you know, two circles are concentric but they are not concentric meaning same center, that's it. But in the typical parlance we consider them to be planar structure. If they are not in the same plane then we don't read them concentric as such as whatever we are, you know, treating them for here. But otherwise there could be circles with the same center. Right? So circles with the same center on the same plane will be called concentric circles, okay? And what is the question? The chord of the larger circle. So this is the outer circle. So AB is the chord. You can see AB is the chord. And it touches the smaller circle. That means AB is tangent to the smaller circle. You have to prove that AC is equal to BC or AC is equal to CB. So how do we prove that? We will use some theorems of our ninth grade which we have studied already. So clearly, so let's say given. Given is AB is AB, no dot here. AB is tangent to smaller circle. Okay? And chord to the bigger circle. Okay? So and what is given? Yeah, so we have to prove. To prove what? That AC is equal to CB where C is point of contact. Contact. Okay? So proof, very simple proof. Not a big deal. So what will you do? You will say since OC, sorry. Since, yeah, so OC is the radius and AB is the tangent to smaller circle, okay? So what does this mean? Circle, sorry, circle. So what does it mean? That OC will be perpendicular to AB and hence the proof. Now we have learned that a perpendicular. Now OC is perpendicular to AB meaning OC is perpendicular cooler to chord AB. Therefore OC will bisect AB. Why? The reason is perpendicular drawn on a chord from the center, from the center bisects the chord. You learned this in the previous grade. So hence done. Okay? So this is done. Okay, so this is another theorem, theorem number five. Okay? Let's go to the next one. The tangent's drawn at the ends of a diameter of a circular parallel. Theorem two, theorem two. Yes, Aditya, here is theorem two. Tell me, repeat, okay? So wait, so let me open this Microsoft. So let me repeat here. So let us say we have a circle, okay? And here is the center, here is the radius and it's given that you have drawn 90 degrees. O, P, Q. Okay, so you have drawn a 90 degree to the radius. You have to prove that PQ is tangent. So I'm saying PQ is not the tangent. Okay, so if PQ is not the tangent, there will be only one tangent passing through the point. So if PQ is not the tangent, then there will be at least a line which is tangent at P, correct? So what is that point? Let's say this is the tangent, right? And I'm saying this is Q dash. So Q dash PQ could be a tangent to the, at this point P is so far so good, Aditya. Is it okay? Yeah, so when that is tangent, then with the previous theorem, we prove that O P Q dash angle is 90 degree by the theorem that tangent and the radius at the point of contact will be perpendicular to each other. So this is O P Q dash is 90 degree. And you're also saying that O P Q is 90 degree. That means Q dash PQ is a straight line. Yes or no? 180 degrees, straight line. So if Q dash PQ is a straight line, that means both these lines, PQ dash and PQ are coincident lines. That means if we have assumed PQ dash to be the tangent, then PQ will be the tangent, right? Is this understood? Else you use this mechanism as well. So let us say you are taking on a point Q. Okay, now clearly, and this is Q dash, clearly O Q is greater than O Q dash. And O Q hence is greater than O P.  That means you can say Q is outside the circle. Right, Q is outside the circle. So if all the points are outside the circle except point P which is lying on the circle, what does it mean? Only P is a common point, common point between the circle and the line. Therefore, the line is tangent. Okay, yes or no? Clear, proof is clear. Shall we proceed? So let's go back to our, yeah, this one. Tens drawn at the ends of a diameter of a circular panel. So it should not be a big deal. What will you do in this case?  So what is given? Oh, this is at the end of the diameter, right? Okay, anyways, so yes, so you can, what is given? Given is CD is a diameter, CD is diameter and E F and E B F, you can write like this also three points. So just to E D F and A C B are tangents at point B and C respectively. Okay, therefore you can say to prove. What is to prove? To prove that E F is parallel to AB, how to prove? Proof, right? So you join this line anyway. So anyways, now you can see O D is perpendicular to E F. And you have to give the region, radius and, radius and tangent at the point of contact, point of contact, radius and tangent at the point of contact are perpendicular to each other, okay? So O D is perpendicular to E F and O C is perpendicular to AB, so you can say angle O D E is equal to 90 degrees and angle O C A is equal to 90 degrees, right? And now either you use interior angle or co-interior angle to be supplementary and things like that, you will get, right? So since angle O D E plus angle O C A is equal to 180 degrees, therefore you can declare what? E D is parallel to AC and since these are lines, so hence you can say E F is parallel to AB. And what is the reason? Co-interior angles on a transversal, on a transversal or supplementary. Okay guys, here, so this is why they will be always parallel. So two tangents drawn and you can see that, not a big deal, so let's say if I draw a line from a, yeah, so if you see, let me draw a line perpendicular to this, yes. So this, yeah, so you can keep rotating point D, sorry, oh, I did not do this part. So if you change D, it is not becoming the tangent, okay, nevermind. So let me prove it. So what you do is draw a tangent from here, from C to this, yes. These are the two tangents, you can see there are two tangents now. And what is that? Let me take this off, so this is not needed. And then, what do I need to do? Construct a parallel line or let's say construct a diameter. So how do you construct a diameter? Join this point, E and construct a line. So E and A, perfect. And now this is the point of contact again. And from here, draw the point of contact drawn and then line perpendicular to this. So perpendicular to this point on this, yeah. So these are the two. So hence what I was trying to tell you is, even if you rotate C, C, it will always be parallel. Correct? See, any case it is going to be parallel only. So till this diameter is there, at the end of the diameter anywhere, you will see parallel tangents always. Okay, done. Next. Line segment joining the points of contact of two parallel tangent circle is diameter of the circle. Yes. So this is the converse of this. So converse of the previous theorem. What is that? The line segment joining the points of contact of two parallel tangent is a diameter of the circle. How do we prove that this is a diameter of the circle? So first of all, they are parallel. So this is given, yeah. Anyone? Anyone would like to prove it? Come on, yeah. Participate. What happened guys? Lack of energy, you're there? Are you there? Or feeling bored? Yes or no? Everybody has gone. Some Indian, Australian cricket match is going on, no? Some match is going on right now? India versus Australia. Who's winning? So you're watching match also. Okay, Aareen is saying, no, what happened to others? There are lots of people who are not communicating at all. Guys, you're there. Let me see your energy. Why? Let me say something. Hello? Yes. You're there? Only Akshita is there. Okay, Riles, Aiden, Nabya, Medha, Hi saying, okay, Moomita is also there. Guys, you're understanding or understanding? Making sense to you or not? Hello people, people, people, people. Yup, so what's the score? Current score? Anyone is watching the India-Australia match? Or is it happening right now? Or no, people are sleeping over there. I don't know. So anyone is following? No one or stumps or already, huh? Who watches tests, right? Test matches the best class of cricket. Okay, it's already, oh, I see. I thought it's going on right now. So old people who retired 10 years ago. Who? Who old people? Oh, their people watch. Hey, no, I also watch test cricket. Yup, so India mostly using difficult position. Oh, is it? Sad. Not like 36 all out, no, like this time around. Or same thing. 36 all out. That was really great. Great day for India. 36 all out. Not like that. Jaideja carried the batting well. Oh, is it? So you are following that means. Yes, good. So fair enough. So come back. So too many injuries. So who got injured, right? Virat Kohli is playing or not? No, is playing or not playing? I have not no clue. Panth got injured, okay? Australia leading by 197, okay? Oh, anyways, Virat Kohli is busy being a dad. Oh my God, this is mean. Aran, why do you say like that? It is, he should be there only, you know? Cricket is important or child is important? Anyway, so what was, yes, how do we prove? Jaganenge, tell me, what do we do? Somebody just said something. Somebody said how to prove. How, what will you do? How will you prove? You will again go for a contradiction maybe. So what you'll do is this. So again, let us say that, let us say that, you know, it's not a diameter. Let us say, let us say that given is, it is given that EF is, AB is parallel to EF and they are tangent. AB and EF are tangent, EF are tangents, okay? This is given, right? Tangents to the circle, okay? And we have to prove, to prove what? DC is the diameter, DC is the diameter. And what is DNC, diameter? So what is DNC? B is DNC are the points of contact, very good. So what will you say? You will start with saying, let it be not, right? So you'll start with what? Proof. You will say, you will say, let DC not be the diameter. Let DC not be the diameter, okay? So that means you must have another center somewhere. Let us say this is the center, some exaggerated view, but let's say this is the center, O dash. So let O dash be the center. Let O dash be the center, be the center. Okay, so clearly what will you say? You will say O dash D and join this, these two, okay? So what can we say now? So therefore, O dash D F is equal to 90 degrees, okay? O dash D F is 90 degrees, we have to write the reason. And then also, but O D F is also 90 degrees, angle, angle. Angle O D F is also 90 degrees, why given, right? It's given, and why is it given?  Yeah, it is given, why is it given? Because you are DC. DC is the line joining the, DC is the shortest distance between E F, isn't it? So you have to prove, you have to, you have to mention that somewhere that DC is the line joining, or DC is the shortest distance between E F. That's what you're saying, okay? Fair enough, that means, this is, these two are possible. When O dash D F is also 90 degree, how can that be possible? This is also 90 degrees, this is also 90 degrees. With the same common arm, two angles are equal, only when, if they are on the opposite side, that means if O dash were here, this side, somewhere here, then you could have said that, okay, these two angles are equal, but O dash and O are on the same side, and there's a common arm to the angle, that means O dash and O must coincide. So O dash and O must coincide. And since O dash was the center, so now you're saying O dash and O are coincident, that means O is the center. Does that make sense? Did you understand? How did we prove that they are, sir, they haven't, they have given DC shortest distance, no. So what you're assuming is, line segment joining the point of contact, point of contact of two parallel, wait a minute, tangent to a circle is diameter to the circle. So hence there are two points of contact, and you are joining them. Now, a line is diameter using linear pair, sorry. Yes, Prishabh is asking a good question. The question is it's not given that this is the shortest distance, is it? So what do we, so hence can we assume that this is how do we prove that that is the shortest distance? So line joining the two, wait a minute, line segment joining the points of contact. Okay, half two parallel, okay, okay, okay, okay. Yes, so how do we see a COD? What are you saying? COD is the line and hence they are tangents. Yes, they are, this is given that they are tangent, but this is fair enough question. How do we prove? Because we have assumed that they are shortest distance. So okay, so the line segment joining the points of contact of two parallel. So they are parallel, okay, fair enough. Yes, look, you will do like this Prishabh. So for all of you, so let's say this is these two are like, these two are parallel lines and this is the transversal, this is the transversal. Okay, the joining point of contact. So now, so these two must be, so what all do we know? So this angle has to be equal to this angle, isn't it? And because of, there is no reason why, you know, this angle is X, then this angle also has to be X. This one, this has to be X, okay. Is my logic correct? If this angle is X, this angle should also be X. Why? Because these are two parallel tangents and there is no reason which tangent to be given as preference. Are you getting the logic? So both the angles must be, the line joining the point of contact must be inclined equally to both the tangents. Am I right? Yes or no? Yes or no? Yes, why? Because, look, this is another way. I'm saying this. Let us say there are two tangents. What did we see? We saw that the two tangents are equally inclined to the line joining the point of, you know, the point from where the tangents are originating at the center. So this angle is, if you remember, gamma is equal to delta. Okay, so if there are two tangents, if there are two tangents, they are parallel to each other, then they are equally inclined to the line joining the point of contact. Am I making sense? Right? So this angle X and this angle is equal to X. They are equally inclined to the, isn't it guys, go low. So they are equal. They are equal, very good. And they are parallel also. The tangents are parallel. So that means two X is equal to 180 degrees. So X is equal to 90 degrees. Got the logic. So if X is 90 degrees, that means this line, DC is the shortest distance anyways. Or even if you don't write that DC is the shortest distance, you can say. Anyways, now you have proved that OD is perpendicular to ED or EF. So either way, your job is done. Yeah, you are. So you have to prove that OD is perpendicular to EF. That's it. So the moment you prove that, you now know that angle ODF is 90 degrees. But you are contradicting yourself when you're saying no angle ODF is equal to 90 degrees because you are not ready to assume or not ready to accept O as the center. So you're saying no, the center lies somewhere else. And that center is ODASH. I'm saying fair enough. That means ODASH DF must be 90. But you just proved this ODF is 90. How is that possible for two angles whose there is only one common arm is there and there are two different points. Is that possible? That both of them are equal without being coincident. So how are equal angles? What are the possibilities? Either angles are like this, X and X. Or if the second arm is on the same side, then there is no other alternative, but the two to be coincident. Is it okay? Yes or no? Did you understand this entire proof? All of you are comfortable? Not comfortable? Tell me any part I'll repeat again from there. Which part is not comfortable? All comfortable? Yes, no. Any part, if you don't want to send a message in public, you can send a message privately to me. I'll not take your name, don't worry. But you please eliminate all your doubts right away. Do not revise this multiple number of times. Only go through this PPT towards the end of your preparation.  It's good enough. Bolo, any doubt, please send me a private note. You can do that in the chat. I will not take your name, don't worry. Okay, so please make sure that you are thorough. Once again, okay. Yes, I will repeat the entire proof. Linear pair way, what is that linear pair thing? You can explain, Meghana, no worries. You can do, if you are comfortable, please do, no problem. Bolo, prove DC or straight line? DC or straight line? Yes, prove it please, how do you prove? Unmute and say. Oh, you are not even sure. Oh, you're not sure. Anyone else wants to prove it? Let's try to prove this. DC is a straight line, yes. How? So let us take this, let us take it here. Control A delete. Okay, so let's try this. Is this the only way to prove or we can do any alternate way? RMTil, it makes sense. Logic has to be correct. So let us say this is a, oh, this circle did not, please, please do linear pair, whichever way you are comfortable. So let us say, this is the tangent. This is the tangent, okay. What is that linear pair thingy doing? Are you saying? So let's see. So this is a line joining the point of contact. Let us say EF, it was E, right? E, so can we draw another line Z over parallel to EF and AB like that? Okay, no problem, EF, AB. And what is that? So draw a line Z over parallel to, okay, just like that. Okay, parallel. Then you can prove through linear pair. How? So let us, this is something EF, GH, let's see. This is O and this was, I think, what was the name? DNC, CD is a straight line, okay. So fair enough. So you are drawing this. So now what is the build of the proof? How do you move? What is the first thing? I drew GH parallel. Now tell me. Constructed, OG parallel to both ends, yes. Now, corresponding angles, action, linear pair theory, the corresponding axle. What is that corresponding action here, angle? Well, oh, so you're assuming this to be a, but this will be 90 degree only when there is, O is the center, but that's not, that's the question. O has to be proven to be center. So we can't start with this 90 degree. So what is the linear pair thing? Please come again. Unmute and say it's better. Unmute. Are in tandem. Yes, sir. Can you hear me? Yes, I can. Go ahead. Sir, we are assuming that O is the center itself. We are not taking that O is not the center. We are trying to prove here that CD is a straight line. And hence it is a diagonal. Yes, CD is a straight line, but then CD is not the line joining the point of contact then, no? I understood.  Okay, so you are saying there are two parallel tangents. Yes. They take the center O and join the center with the point of contact and then draw a parallel GH, a line parallel, line GH parallel to EF and AB, fair enough. But in that case, do not say that CD is the line joining the point of contact. Okay, then eventually you prove that. Yes, sir. So hence you will say that, okay, so hence, yes, then it follows correctly. So OD is perpendicular to what, DF? Yes, sir. No, sorry. OD is perpendicular to DF. And also, since OD is perpendicular to DF and DF is parallel to OG or OH. Therefore, OD is also perpendicular to DF. I'm sorry, what's the other one? OH. OD is also perpendicular to OH. So this is 90 degrees. Similarly, similarly, you can prove similarly, OC is perpendicular to CB and CB is parallel to OH. Therefore, OC has to be perpendicular to CB. Sorry, OH. OH. So this is, so you can now say that angle DOH plus angle COH is equal to 90 plus 90, that is 180 degrees. Hence, hence, DC is a straight line. DC is a straight, straight line. Now you have to say, but DC is a diameter which is also the line joining the two points of contact. Okay, so hence proved. You didn't stand. So initially we cannot start with DC to be a straight line. So forget about DC. So eventually you prove that DC is coming out to be a straight line. In my proof, I started with the assumption that DC is a straight line. And since there are two parallel tangents and DNC are on the opposite sides of the, let's say the two parallel tangents will always be equally inclined to the point of contact. So hence, I consider that to be the shorter distance or already proved that that is 90 degrees then to contradiction. Whichever you can remember, you're comfortable, please do that. Now, angle between the two tangents from an external point to a circle is supplementary to the angle subtended by the line segments joining the point of contact to the center. What does it mean? It means that, now let us quickly prove this. You'll mean that this angle here, this alpha plus beta is 180 degrees. So prove that alpha plus beta is 180 degrees. That is what they are saying. What are they saying? Angle between the two tangents, right? Angle between the two tangents drawn from an external point to a circle is supplementary. So beta is the first angle they are talking about. Is supplementary to the angle subtended by the line segments joining the point of contact to the center. So this is the line segment joining the point of contact and the angle subtended by AB on O is alpha. So you have to prove that alpha beta is 180 degrees. Very easy. Yes, someone is saying Aditya is saying angle sum property, very good. So you already know that angle A plus angle B plus angle or I have mentioned alpha. So alpha plus beta is 360 degree. But you already know that angle A is 90 degrees. Angle B is 90 degrees. So hence this 90 and this 90. So alpha plus beta will be 360 degrees minus angle A minus angle B, which is 180 degrees. Right? Since angle A is equal to angle B is equal to 90 degrees. Okay, so this should not be a big deal. Let's go to the next one. There is one and only one tangent at any point on the circumference of a circle. How do we prove it? Again, let there be two points or let there be two tangents from there you can start. Or there is one and only one tangent at any point on the circumference of the circle. What will you do here? So you can go through a contradiction route. Then let there be two tangents. Okay, so let's say PQ is one tangent. Is this similar to theorem two? Yes, somewhat. Yes, so you can say that. So what I'm saying is, yeah. So let there be two tangents. We've got all PQ and let's say another one, PQ dash. Let there be two. So if there are two tangent tangents, then we know that angle again, same thing, OPQ is equal to 90 degrees and similarly OPQ dash will also be 90 degrees. But this will contradict again. Why? This is possible only when Q and Q dash are coincident. Did you understand? Right? No contradiction, by contradiction you can prove. So far so good, guys, any difficulty. So we're quickly covering the theorems. Last one, the perpendicular at the point of contact of a tangent to a circle passes through the center. Okay, so again, let it not pass through the center. Very easy, these are very easy proves. Perpendicular at the point of contact of the tangent to a circle pass through the center. So again, let it not pass through the center. Okay, so let's say if it is not passing through the center, let us say yeah, so then you join the center and point P. Yeah, join the center and let's say, then here we are getting QPT is 90 degree. Why? This is given that the perpendicular at the point of contact of the, we are already assuming that let QP be perpendicular to PP. Okay, but we are saying QP is not containing the center. But QP doesn't, sorry, QP doesn't carry or pass through the center. Okay, so let POR be the diameter. The diameter, so clearly, clearly what? O is the center, so that means angle OPT must be equal to 90 degrees by the theorem. But you are saying QPT is 90 degree by given, right? OPT is 90 degree, that means angle RPT, RPT is 90 degrees. And you're also saying angle QPT is 90 degrees. Both are possible when, when are these two possible? When R and Q coincident, right? Here, all theorems, these are the set of 10 theorems which are there in the chapter of circles. Let's, let's now go to question. Easy one, application, one marker. Usually the questions are definitely one marker, one plus three or two plus two. Not more than four marks question will be there. There were cases where, or exams or years where there were five marks questions as well. Yes, 56 centimeter is the right answer, right? So clearly PQ is equal to PT is equal to 28 centimeter. And you have to mention the reason, tangents on a circle from same external point are equal. Okay. And now perimeter, perimeter is PL plus LM plus MP so PL plus LM can be, sorry, this is LM. LM can be written as LM plus NM plus MP. So this is equal to PL plus LQ, isn't it? So L is the external point from where two tangents are drawn. So two tangents are drawn. So like this, L, L, so they're equal. PL plus LQ plus NM, colloquial, MT plus MP and club them together, put the brackets also to emphasize. So PL plus LQ is PQ plus PT is equal to two PQ is equal to 56 centimeter. Good, one marks with you. So one done. This is question for, so total two. If two tangents are inclined at 60 degree, two tangents are inclined at 60 degree are drawn to a circle of radius three centimeters and find the length of each tangent. So quickly draw the diagram. Do not forget to draw diagrams, okay? Two tangents are inclined at 60 degree. So each one of them are inclined then at 30, 30 degrees. So this is 30, this is 30, isn't it? So radius three centimeter. What is radius? This is three. So this is 90 degrees. So you're used to geometry. Find the length of each tangent. So this is length L, right? So about three upon L is tan 30, no? Three upon L is tan 30, yes or no? Tan 30. So three upon L is equal to one upon root three. So L is three root three centimeter. All of you getting it? Three root three, okay? Trignometry, fair enough. One, one, two marks done. This is, this was in R. So that's a total of three. PQ is a tangent to a circle with center O at point P. If triangle OPQ is an isosceles triangle, then find OQP. So clearly PQ is a tangent to a circle with center O. So quickly a circle O. PQ is a tangent to a circle. This is O at point P. If triangle OPQ, OPQ is an isosceles triangle, then find OQP. Oh yeah, this is very easy. OQP, what is this? 45 degrees, isn't it? So it's given that PQ is tangent. So PQ and triangle OPQ is isosceles. Isosceles only cases this will be equal to this. There is absolutely no other way because OQ is going to be hypotenuse and this is 90. So fair enough, you're 90. So this angle is 45 degrees. This is also 45 degrees. Okay, clear. So three marks in your kitty. Next, all of you are okay. Am I going fast? Guys, tell me, are you able to cope up quick? Any doubt, please keep sending your doubts to me privately also. No need to worry that, okay, I am lying behind. You can always go back and watch this video. There are lots of questions, 20 plus questions. So let's try to solve them all. Can you slow down a bit? Okay, okay, slow it slowing down with the next question. Do it. In the figure quadrilateral AVCD is a circumscribing is circumscribing a circle with center O and AD is perpendicular to AB. Where is AB by the way? This is B. If radius of in circle is 10 centimeter, then the value of X is, find out the value of X guys. Aditi, no, everyone is saying 21. Okay, Y 21, Y 21. Okay, anyways, it's easy. Why, because, there you go. This will be equal to 27. Why? Equal tangents are equal from the same point. So C is the point from where two tangents are drawn. So CR is equal to CQ is equal to 27 centimeter. So obviously BQ will be equal to 38 minus 27, which is 11 centimeter. So BQ is 11. So if BQ is 11, then my dear friend QP is also 11. Same logic, everywhere same. Tangents drawn to a circle from the same point are equal. So BP, sorry, BQ name, BQ, oh, sorry, BP, BP. BP is equal to BQ, yes. BP is equal to BQ, 11. So what is AB? X is equal to AP plus PB or BP. Now clearly AP is 10. Why? Complete the square. So AP is 10. So 10 plus 11, 21 centimeter. But you have to also prove that AP is 10. So don't really write it, you have to prove why AP is 10. Why is AP 10? Can anyone tell me why is AP 10? So you basically complete the square, drop a perpendicular from, or join OP. So OPAS is a square. Why is it a square? Why is it a square for that matter? Because OPAS may AP is equal to AS and it's a rectangle. It is a rectangle. Why it is a rectangle? Because there are so many 90 degrees rectangle. Why? Because angle A is equal to 90 degrees and angle OP, A is equal to 90 degrees and angle OS, A is equal to 90 degrees. So it's a rectangle with AP is equal to AS. That means, therefore, OPAS is a square. Hence that is 10. Clear? Very good. So Aditi, next. This is one mark again. Actual board pepper, 20, 20. BC is equal to, all of you are getting 10 centimeters. This is one marker, so it should not be a big deal. Again, so what is this? Clearly, AP is equal to AR, is equal to four centimeter. That means RC is equal to AC minus. So this is one marker, but still let's say it comes for two markers and like these steps. So RC is equal to AC minus AR is equal to 11 centimeter minus four, that is seven centimeter. That means RC is equal to QC is equal to seven centimeter. Similarly, BQ or BQ, BP, BP is equal to BQ, is equal to three centimeter. Now BC is equal to BQ plus QC is equal to three plus seven. Now is the pace okay or I'm still going fast? Tell me guys, shall I slow down more? You want some more time? Yes or no? Is it fine with all or you want me to slow down further? No problem. Okay, so let's do it. This is two marks. Two tangents TP and DQ are drawn to a circle with center O from an external prove that PTQ is twice OPQ. Proving question, two marks. So to write given, proof, construction, all that stuff. Just write done when done. Done, okay. Very good. This particular question or the theme of this question you are going to use a lot in physics this year that coming year. So Prishan, Srishti, Sai, Rishita, Saptam, all of you have done, very good. Yes, you can use ASP also, I think. Done, Aniruddha Sai done, Siddharth Kartik done, Alisha Prakash done, very good. Arjun, Akshita done, very good. Come on, come on, fellas, chalo chalo, done, not done. You can send me, if you don't want to post your listing in the group, send me at least privately so that I can understand whether you are with the pace. So please make sure that you respond back to me. Okay, guys, done or not done, still struggling. Join TO, join OQ also, that's done. So we'll see. Okay, so to PTQ, no, ha, defo. Now, PTQ in triangle PTQ, PT is equal to PTQ. Hence, angle TPQ is equal to angle TQP, correct? Is equal to long method, why? This is equal to half of 180 degrees minus angle PTQ. Yes or no, guys, is that okay? Is that okay, angle TPQ is equal to angle, yep. Okay, now, come to triangle OPQ. Since OP is equal to OQ, there are radius. Therefore, angle OPQ is equal to angle OQP is equal to 180 degrees minus angle POQ by two. Okay, yes or no? Bolo dosto, right? So from here, there are two, three methods you can follow. Okay, or else what you can otherwise do is, you can also write this as, instead of this, you can write this as OPQ is 90 degrees minus angle OQPT. Angle QPT, is it okay? There we go, once again, I'm writing. What I'm writing, just pay attention. This is 90 degree, guys. So angle OPQ, OPQ is equal to 90 degrees minus angle QPT, right or wrong, right or wrong? Correct, yes or no? Correct or right? All of you are with me on this. Okay, next, angle OQP is equal to 90 degrees minus angle PQT or PQT, right? Is that, now, add both of them. So OPQ plus O, OPQ plus OQP, can it be written as two times angle OPQ, LHS if you add, both are equal, so I can add that. And here it is 180 degrees minus 180 degrees minus two times triangle, two times triangle QPT, angle QPT, right? Or let it be like this, angle is only, this will help PQT. So 180 degrees minus QPT, TPQ is nothing but angle PTQ, right? Did you understand? So I wrote this, I wrote this added, I got 90 plus 90, 180 minus QPT and minus PQT.  So angle PTQ, understood, not understood. Send me each one of you, please respond, respond. So that, you know, things should be clear, okay? Clear, clear to everyone. So this was two marker, okay? Next, next, do this, easy, one mark. I thought I was throwing this simple. PB is equal to, this is like you have to do mentally. PB is equal to, yep, good, mentally. So AP is clearly four, why four? Because this is 90 degree, why 90 degree? Because OP is perpendicular to AB, why? Because at the point of contact, radius and tangent are perpendicular, okay? So hence this is four and this also will be four, why? Because previous question once, okay, I'll show you. So is this clear to everyone? This question is four centimeter. Any doubt in this? This one, I will go to the previous one, don't worry. Is this clear to everyone for why it is four? OP bisects AB, we learned in grade nine that a chord is bisected by the perpendicular dropped from the center onto it, okay? So OP is perpendicular to AB, hence clearly, AP is equal to PB from nine grade stuff and AP is equal to four. So PB is four, okay, going to previous one. Once again, let me clear this off, wait.  Okay, let me use this space. Oh no, wait, when I'm spending time, I should do it properly. Unfortunately, the presentation PowerPoints do not have that good scribbling mechanism, so hence becomes difficult at times, but nevermind. Okay, so I have cleared some space now. Once again, I have joined OQ. Now I know that this is 90 degrees, this is 90 degrees. So I'm writing angle OPQ is equal to 90 degrees minus angle QPT, this is clear. I hope this is clear, QPT observed and you will get it, QPT, right? Angle, similarly, angle OQP, OQP is equal to 90 degrees minus angle PQT, okay? Now I simply added them. So OPQ and QOPQ plus OQP is equal to 90 plus 90 minus angle QPT plus PQT, okay? Now OPQ and OQP, both are equal, right? So hence I can write this as two angle OPQ, why? Because OP is equal to OQ, hence angle OPQ will be equal to angle OQP, okay? Now in the right hand side, I have 180 degrees minus angle QPT, so observe QPT, where is QPT? This and where is PQT? This, so this you are adding and subtracting from 180 degrees, don't you think that is nothing but this angle, so right, 180 minus PQT, right? So you can write this since angle QPT plus angle PQT plus angle T let's say is 180 degrees. So angle T will be 180 degrees minus angle QPT plus angle PQT, right? So instead of this here, I'm writing angle T and angle T is PT, okay? Good, next, this one, three marks. So there was no diagram given for this figure, so how do you imagine how this will be done, okay? I hope you are able to draw the diagram into people three, okay? Think about it, the circle touches the side BCF triangle. So the circle is outside the triangle because it is also touching extended sides. Did you get the diagram first? So first draw a triangle ABC, A, B, C. And they're saying there's a circle which is touching what? Side BCF triangle, so there are two possibilities, either the circle is like this or the circle is like that. Now if in the first case, if you will not touch the extended sides only in the second case it will touch the extended sides, A, B and BC. So the circle is something like this. This will be the circle, this will be the diagram. There is some center here, okay? So let us say this point is B, this is C. Let us say this is what is this point? This is P and A, B extended is Q and R. This is a diagram, if you've drawn the diagram correctly you'll be able to do it. So AQ, right? So AQ is equal to AR, clearly. Why? Tangents drawn from the same point, from an external point onto a circle are equal. Similarly, if you see BQ is equal to BP and PC is equal to CR. Using these three you can prove AQ. So if you start from RHS, AB plus BC plus CA, half is equal to half of AB plus BC can be written as BP plus PC plus CA. This is half AB plus BP. Now BP can be written as BQ and PC can be written as CR plus CA. Similar question we just did above where there was a numerical based question. Now if you see this is half AB plus BQ is AQ and this one AC plus CR is AR. But AQ, AR are themselves same, so half into twice AQ. So hence AQ, okay? So what was the catch of the problem first to draw the diagram? So let's say when you were drawing the diagram, let's say you are thinking you're not able to crack it in the first instance.  So hence the circle has to be on the other side, okay? And we draw the diagram and then the moment you got this, you know the technique now. So there are tangents, multiple equal tangents are being seen. So that should give you a hint that you have to approach like that. So AB and we saw a problem over in the previous slides also. So use the only one underlying concept that tangents from the point to a circle are equal. And instead of starting from here to go here you start from here and go there. And it was easy. I hope, is this question understood? So please try to map it in your brain right now, okay? So when you do multiple number of times you will be familiar with these kinds of questions as well. One mark should not be much of a time consuming question. Exactly same thing we did previously just now. It proved 25 degrees, very good. So easy, very easy. So this is X. This is also X. So X is 180 degrees minus 50 by two which is 65 degrees. And angle OAP is 90 degrees because OAP will be perpendicular to OA is perpendicular to PA. So angle OAB is equal to 90 degrees minus 65 degrees. 25 degrees. This one. So be careful which angle is 60 degrees here. QPT. You have to find out P, R, Q. No, no, no, nothing like that. You don't need to. In one marker two, three steps over. Question should be over. Maximum three steps. Everyone is saying 120, 120, 120, 120. How, how 120? Who will explain? 120, Shobita. Sure. So hence this, okay, let me. Oh, wow. What? How do I? So this is a question. And if you see this angle is how much? How much is this angle? The red one. So this angle is 30 degrees. Right? Angle PRQ, you're getting 180 degrees, actually. Why? This is 30. This is also 30. This is 120. So this is 240. So this is 120. PRQ, you know, 120. All right. Everyone got it? Yes or no? This is how we have to do it, right? So once again, 30, 30. So by angles on property, this angle is 120. So the reflex angle is 240. Now this arc, bigger arc, the larger arc, PQ is subtending an angle 240 at the center. So at any other point in the segment, in the same segment, it will be 120, half of that. So 120. Clear? Okay. Next. In the figure four, AB is a chord of a circle with center O. AOC is the diameter. And AT is tangent. Prove that back is equal to ACB. Could you go back? Yes, I can go back when, yes. Previous question. Yes, once again, they go. Let me delete. Once again. See, this angle is 60 degrees. This is 30 degrees. I hope this is clear because 90 minus 60. Okay, so hence this is also 30. Isosceles triangle, same radius. So 30, 30. Now, if that is 30, 30, then this angle is 120. Without doubt. Angle some property. Okay. If that is 120, then the reflex angle is, this angle is 240. Isn't there another chord? Subtracts double the angle at the center compared to, that's what we are using. So hence, they go. So what is the arc? QP is the arc. Major arc, which is subtending 240 degrees. What is the impression? Can you unmute and say, I didn't understand your question. That's what you mean? There's a theorem that says that the chord, which is PQ over here, subtends double the angle at the center, then it does at any other point in the circle. And the same segment. So then. Same segment. So you're right, but it has to be here. Okay. So if this is 120, this will be 60 degrees. And exactly same thing I'm using. So I'm exactly same thing, but this time around, this angle PRQ is on the opposite segment. Can you see that? It is not in the same segment. Same segment as in, see for this, this angle PRQ is the, so the double of this is angle QOP reflex. Getting it? I'm using the same thing, but you are picking the wrong angle. Once again, they go. Let me just take it away. Okay. So I'm using the same theorem, but you have to pick the right angles. Okay. So if this is 120, then this angle is going to be 60 degrees, half of it. But this angle is, you know, so this is half of the angle subtended by larger arc. QP at the center. And how much is that angle 240 degrees? And hence now it is subtending this angle at the segment opposite segment. So hence this is 120 degrees. Okay. Half of 240, clear? Next, this one. I use alternate segment theorem. What is that? Aditya, alternate segment theorem. Can you unmute and say? This question is done. This is easy. So how to do this one? So the diameter, AC is diameter. So clearly this is 90 degree. Yes or no. And if this angle is X and this angle is 90 degree as well. So angle, this angle is 90 degree minus X and this angle is also 90 degree minus X. Clear? So angle, mat is equal to ACV. Did you understand all of you? What did I do? Since AC is a diameter, so ABC has to be 90 degrees and consider this to be X. And clearly OAT is also 90 degrees. So this angle is X if then this angle is 90 degree minus X. And in the triangle ABC by angle some property AC has to be 90 degree minus X and hence they are equal. Both are 90 degree minus X. Shall I proceed? Understood by all? Go ahead. This we have already done so we'll skip it. This is the theorem, but this was asked in 1920. Board paper. A tangent to a circle is perpendicular to the radius to the point of contact. The first theorem I told you. So we'll not do it. Same thing, prove that the angle between the two tangent drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the point of contact at the center. Isn't it? This was a theorem again. We had done this. Which one? Prove that. This is the, these are the two tangents. Sorry. This one. And this one. And they're saying. Prove that the angle between the two tangents, beta. Drawn from an external point P. Let's say this is Q. This is R. This is O. Right. Is supplementary to the angle subtended by the line segment joining the point of contact. So what is the line? This angle alpha. And we have to prove alpha plus beta is equal to 180 degrees. Yes. So we did this angle some property of a quadrilateral because this is 90. This is 90. So alpha plus beta will have to be 180 degrees. So we had done this previously. So not doing this again. This one. PTQ. Very easy. Application. Angle between tangent and chord. Angle made by the chord in the alternate segment. Oh, like that. Yes. But then there was, it was not that case. Just a minute. Let me see. Just a minute folks. You can. Yes. Yes, you can. Mr. So you take the supplementary, the other angle. Other angle. So in question 21. 21. Oh, yes. So what did you take the angle which is 180 minus angle QPT. QPT 180 minus QPT. Okay. And then which is 120 degree. And then. Actually, it is equal to angle PRQ. You use it directly. Now you can use that. But yeah, if you want one more case, you can use it directly also. Yes, you can use that. So what Aditya is using is this. So let us say you have a tangent and you have a chord. So angle made by the chord, which angle, this angle. X is equal to this X. That's what you're using, right? Yes, sir. Yeah. So you angle made by a tangent and a chord is equal to angle subtended by the chord in the opposite segment. So you can use that also directly. Yes, it's okay. You know the proof, right? Yes. How do you prove that this angle is equal to this angle? X. So you construct the radius. This one? Yes, sir. And the other angle is 90 minus X. The other angle is 90 minus X, yes. And so since from the triangle. This angle is 90 minus X. Yes, sir. 90 minus X. This is 90 minus X. 90 minus X radius is equal. So this angle will be 180 minus, yes. So this is 2X. So this is X, correct? Very good. You can use it directly, no problem. Good. So yes, this is done. This will be 150, 180 minus 115. No questions. Isn't it? 180 minus 115. Correct? So 65 degrees, cool. Very good. Next, D. This one is easy. From an external point Q, length of the tangent to a circle is 5. And the distance of Q from the center is 8. Radius of the circle is, draw the diagram quickly. Circle, here is Q, tangent. This is 5. Distance of Q from the center is 8, 8. The radius of the circle, Pythagoras theorem, 64 minus 25, 39 under root, right? So let's say R. So R square plus 5 square is equal to 8 square, OK? Next, CD, length of CD, 4 centimeter, yes. Why will that be? Understood? So AC, CB is 4. And CD is equal to 4. CD is equal to CB, right? This is 2 marks. Hello, OK, I doubt. Once again, AC is equal to CD. And CD is equal to BC. So BC will be equal to AC. And hence, 4, 4 each. Shall I give you a hint? Join OC. Yes. What happened, guys? Anyone else? No? Arjun and Arin. So you have to join OC. And when you join OC, then this angle is equal to this angle. And this angle is equal to this angle. Yes or no? So hence, 2x plus 2y is 180 degrees. So x plus y is 180 degrees, 190 degrees, right? Once again. Everybody understood this part? Y is x equals 2x there because the two triangles are congruent. How? RHS. Which two triangles POA is congruent to OCA? They are congruent. Why? There is a common side. There is a 90 degree, both, and radius. Or either you can say AP is equal to AC or SSS, whichever way. There are multiple ways you can prove they are equal. So once again, this was the case. This is the line. This is the center. This is the tangent. And I had joined that. I had joined that. I am saying PA is equal to CA. These two sides are equals. AO is equal to AO, common. And these are the radius. You can use SSS also. So the moment they are congruent, this is x. This is x. By the same logic, that will be y and y. So hence, 2x plus 2y is 180 degrees. So x plus y is 90 degrees. And that is what you want. AOB is x plus y. Do we need to prove that POQ is a straight line? POQ. Two parallel times is no. You don't need to. Why would you? I did not use anyways. Yes, POQ. Sorry, yes. POQ's straight line is, no, no, we had a theorem. So you have to write that. That the point of contact. What was the theorem? If there are two parallel tangents, the point of contact will be diameter. Sorry, the line joining the point of contact will pass through the center. We dealt with theorem number 9 above, 9 or 10, something like that, above. It's not an NCRD theorem, OK? It's not an NCRD theorem. But I don't really think that you can use it directly. But because if you go for proving this, it will become too huge a proof. OK, so wait. Wait a minute. But O is anyways given, center, omita. O is given here to be center. So OP and OQ, that means POQ is a diameter. Because 90, 90 degrees, is it it? Did you understand? So you have to, yeah, if you want to prove it, you can always prove that PQ is a diameter. PQ is a diameter, why? Why is PQ a diameter? Because it is passing through O first of all. Angle OPX is equal to 90 degrees. And angle OPOQM is also 90 degrees. And PX and QM are parallel. That means PQ is a straight line, right? So PX and QM are parallel. So you are saying OP is perpendicular to PX. And OQ is perpendicular to QM. And PX is parallel to QM. Therefore, OP is either parallel to OQ or OPQ is a straight line. So you can first prove that. But these are for three marks, so I would have assumed this theorem to be true. And then there are two congruences to be proven. And then, right? But if you have time, please. So in such questions, let's say if you have a doubt. So keep that in your mind. Put a star mark maybe somewhere, wherever you can. Or write a question number 30 to be revisited, something like that. And then if you have time, then add that proof. Append that proof in the bottom of that. Leave some space. Is that understood? So leave some space. And you think that, OK, if time permits, I'll come back and append that proof with this. So why is PQ a diameter? You can append it towards the end. OK, Moomin, that's clear. So if you think that, OK, I might need to prove it once again. Or some bit more. So just keep some space over there. Come back if you have time. Any ways you will be left with some time and quickly add that proof. Fair enough. So there were a few more similar type of questions here. So this is another proof. ABCD is drawn to circumscribe a circle. So you have to prove that AB plus CD is equal to BC plus AD. This will be also easy. So AB plus CD. So you have to break them into the parts. So let's say this is X. This is X. This is Y. This is Y. This is U. This is U. And this is V. And this is V. So you can clearly see AB is equal to X plus U. And CD. CD is V plus Y. So this X of this can be replaced by this X. And this Y can be replaced by this Y. So you'll get AD on the other side. And similarly, U plus V will be BC. I just gave you a quick approach of what to do. So AB plus CD, you can see. AB plus CD. Let me prove it only.  Now rearrange them.  So X plus Y is now AD. And this is BC. So done. So AB plus CD is equal to, right? Like that, this is easy. This one's similar. Similar. Again, it was repeated. See, not many times it repeats. Similar. We just proved this one also. This is, see, this question was in 18, 19, year before. Year before. Similar type of questions. This one also, we just did it. All are similar type of questions. This one proved that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle. Did you get this question? Question is this. Prove that just two minutes and I'll wrap it up. Prove that opposite sides of a quadrilateral circumscribing. This is a cyclic quadrangle, basically. Prove that opposite sides of a quadrilateral circumscribing a circle. Oh, circumscribing. Sorry, this is inscribing. This is wrong. No. Circumsccribing a circle is this. So circle will be within the quadrangle. So something like that. OK. So A, B, C, and D. OK. And they're saying subtend supplement opposite sides of a quadrilateral. So this is opposite side. This is the center. So what are they saying? The question is prove that x plus y is 180 degrees. x plus y is 180 degrees. Can you prove that? It will be easy if you really look at it. Why? Because this angle 1 is equal to angle 1. This angle 2 is equal to angle 2. Angle 3 is equal to angle 3. And angle 4 is equal to angle 4. Why are they equal? Because we had a theorem saying the line, the two tangents are equally inclined to the line joining the center and the point from where the tangents were drawn. Am I right when I write like this? So what is x plus y?  Correct? And what is y? 180 degrees minus angle 3 plus angle 4. And G. Yeah. So this is 360 degrees minus angle 1 plus angle 2 plus angle 3 plus angle 4, which is nothing but if I multiply by 2 and divide by 2, right?  Cyclic, it's a quadrilateral, so 360 degrees. So this is 360 degrees minus 360 degrees by 2, which is 180. Got this proof? What did I do? I wrote x first. I wrote y. Then I added.   Angle some property of a quadrilateral. So hence I replace this by 180 degrees, isn't it?  So 360 degrees minus this 180 degrees. Is this proof clear?  Do once again, as I already gave you the direction. You can do that. Solve these. These are the, see how beautifully people have written. So this you can take as a reference. How to prove is written. They have not given the given part of it. Construction is written. So elaborately to prove. So you can start from given instead of writing all of this. Start from given, then write construction, then to prove. And we know that and all that step by step, step by step. Beautiful presentation. This is for one marker. So rough diagram is drawn. You don't need to take a compass and start drawing there over, you know, perfect circle and all that. What are you doing there? Is it OK? So this is for one mark. Fair enough. So I'm sorry, I took a little bit more time today. So OK. So we'll stop here. In the next class, we will carry on with whatever is a topic assigned to that date. Maybe we'll constructions, we will have a different date altogether. Is it OK? So please keep solving. Tomorrow again, one set of mock papers will be released. You can solve them at home. OK. Thank you. See you again. Bye-bye. Take care.