Question

Find the equation to that chord of the circle $x^2+y^2=81$ which is bisected at the point $(-2,3)$, and its pole with respect to the circle.

Answer 1

Let the chord be $y=mx+c$ ... $\left(i\right)$

The line joining the midpoint of the chord to centre of the circle is

perpendicular to the circle, whose slope will be $-\frac{1}{m}$

$\frac{-1}{m}=\frac{3-0}{-2-0}$

$\Rightarrow m=\frac{2}{3}$

To calculate $c,$ put value $(-2,3)$ in the equation $\left(i\right)$ of line

$3-\frac{2}{3}\times (-2)=c$

$\Rightarrow c=\frac{13}{3}$

Thus, the chord is given by $3y-2x=13$

Let the pole of this chord be $(h,k)$

Polar of $(h,k)$ is given by $xh+yk=81$

Since the chord and polar are same line, there equation must be identical

$\Rightarrow \frac{h}{-2}=\frac{k}{3}=\frac{81}{13}$

$\Rightarrow (h,k)=(\frac{-162}{13},\frac{243}{13})$