Question

Tangents are drawn to the circle $x^2+y^2=25$ from the point $(13,0)$. Prove that the angle between them is $2{\tan }^{-1}\left(5/12\right)$ and their equations are $12y+5x+65=0$ and $12y-5x-65=0$.

Answer 1

Any line through $(13,0)$ is
$y-0=m(x-13).....\left(1\right)$
or $mx-y-13m=0$
The condition of tangency $p=r$ gives
$\frac{-13m}{\sqrt{(m^2+1)}}=5$ or $169m^2=25m^2+25$
$\therefore m=\pm 5/12$.
Hence we get the equations of tangents from $\left(1\right)$ on putting the values of $m$ as given.
The two tangents are equally inclined to the axes and hence the angle between them is $2\theta$ where $\tan \theta =5/12$. $\therefore 2\theta =2{\tan }^{-1}\left(5/12\right)$
Alt. The two tangents will be equally inclined to the line of centres. If $t$ be the length of tangent, then $t^2=S^{\prime }=144.\therefore t=12$
Also radius is $5$

$\therefore \tan \theta =5/12$
Hence $2\theta =2{\tan }^{-1}\left(5/12\right)$.