Question

The pole of the chord of the circle $x^2+y^2=81$, the chord being bisected at $(-2,3)$ is

• A. $\left(\frac{-162}{13}\frac{243}{13}\right)$
• B. $\left(\frac{162}{13},\frac{-243}{13}\right)$
• C. $\left(\frac{-162}{13},\frac{-243}{13}\right)$
• D. $\left(\frac{162}{13},\frac{243}{13}\right)$

Answer 1

Answer: (B)

$\left(\frac{162}{13},\frac{-243}{13}\right)$

let $Q(-2,3)$ is a mid-point of polar of $O$ at $x^2+y^2=81$
$\therefore OP\times OQ=r^2$
$OP\sqrt{13}=81$
$OP=\frac{81}{\sqrt{13}}$
So, equation of $OQ$ is
$y=\frac{-3}{2}x$
Let, $P(h,\frac{-3}{2}h)$
$\therefore OP=\frac{81}{\sqrt{13}}$
$h^2+\frac{9}{4}{h}^2=\frac{(81{)}^2}{13}$
$h^2=4\frac{(81{)}^2}{(13{)}^2}$
$h=\frac{2\times 81}{13}=\frac{162}{13}$
$\therefore P(\frac{162}{13},\frac{-243}{13})$