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Find the equations of the two lines which pass through the point $(0,4)$ and form tangents to a circle of radius $2$, centered on the origin.

Firstly, we have the equation of the circle $x^{2}+y^{2}=4$. Which we can rearrange to get $y$ in terms of $x$:

$$y^{2}=4-x^{2}\implies y=\pm\sqrt{4-x^{2}}$$

However, we know that the tangents must touch points on the top half of the circle, therefore we can simply take the principle square root, $y=\sqrt{4-x^{2}}$. Moreover, since the lines originate from the same point, and are tangential to the same circle, the two tangential points are $(x,\sqrt{4-x^{2}})$ and $(-x,\sqrt{4-x^{2}})$.

The gradients of the tangents at these points can be found by implictly differentiating the original equation and obtaining:


However, I'm unsure how to go about completing this problem. I know it's a simple question, but I simply cannot see how to solve it.

Thanks in advance!

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This is the exact copy of your question on Yahoo! Answers –  user31280 Oct 29 '12 at 12:39
@F'OlaYinka That's not my question, I was not even aware of its existence. –  Shaktal Oct 29 '12 at 12:39
I'm not saying that you posted it there, I'm only saying that it's a copy of yours. –  user31280 Oct 29 '12 at 12:41
The answer given on the Yahoo! Answers page is not correct. –  Peter Phipps Oct 29 '12 at 13:07

2 Answers 2

up vote 1 down vote accepted

Call $\,P=(0,4)\,$ , and let $\,A:=(a,b)\,$ be one of the two tangency points. Since the tangent line's perpendicular to the circle's radius at the point of tangency, we get:

$$m_{AP}=-\frac{b-4}{a}\,\,,\,m_A=\frac{b}{a}\Longrightarrow \frac{b}{a}=\frac{a}{b-4}\Longrightarrow a^2=b^2-4b$$

But we also have $\,a^2+b^2=4\Longrightarrow a^2=4-b^2\,$ , since $\,A\,$ belongs to the circle, so we get the quadratic

$$a^2=4-b^2=b^2-4b\Longrightarrow b^2-2b-2=0\Longrightarrow b_{1,2}=1\pm\sqrt2$$

and now find out the $\,a's\,$ and etc...and without calculus!

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Thank you! I knew it was a simple problem, but for some reason I had a complete mind blank! Thanks so much! :) –  Shaktal Oct 29 '12 at 12:33
Any time........ –  DonAntonio Oct 29 '12 at 12:39

The equation of any line passing through $(0,4),$ can be written as $$\frac{y-4}{x-0}=m$$ (where $m$ is the gradient), or $y=mx+4$

Let this line touches the given circle $x^2+y^2=4$ at $(h,k),$

so, $k=mh+4$ and $h^2+k^2=4-->(i)$

Putting the value of $k$ in $(i),$

we get $h^2+(mh+4)^2=4$ or $(1+m^2)h^2+8hm+12=0$

The $2$ roots of this quadratic equation represents the points of contact. For tangency, both the roots will be same, i.e., the discriminant must be $0$.

$(-8m)^2=4(1+m^2)12\implies m=\pm \sqrt 3$

So, the equation of the two lines are $$\frac{y-4}x=m=\pm \sqrt 3$$

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