Question

Find the pole of $x+y-2=0$ with respect to the circle $x^2+y^2-4x+6y-12=0$.

Answer 1

Given equation of polar-

$x+y-2=0.....\left(1\right)$

Let $P(h,k)$ be the pole of line $x+y=2$ w.r.t the circle $x^2+y^2-4x+6y-12=0$

Therefore the polar of $P$ w.r.t. the circle is-

$hx+ky-2(x+h)+3(y+k)-12=0$

$(h-2)x+(k+3)y-2h+3k-12=0.....\left(2\right)$

Now equation $\left(1\right)\&\left(2\right)$ are same.

Therefore,

$\frac{h-2}{1}=\frac{k+3}{1}=\frac{-2h+3k-12}{2}$

$\Rightarrow 4h-3k+8=0.....\left(3\right)$

$\Rightarrow 2h-k+18=0.....\left(4\right)$

Multiplying equation $\left(4\right)$ by $3$, we get

$6h-3k+54=0.....\left(5\right)$

Subtracting equation $\left(3\right)$ from $\left(5\right)$, we have

$(6h-3k+54)-(4h-3k+8)=0$

$\Rightarrow 2h+46=0$

$\Rightarrow h=-23$

Substituting the value of $h$ in equation $\left(4\right)$, we have

$-46-k+18=0$

$k=-28$

Hence the pole of line $x+y+2=0$ w.r.t. the circle $x^2+y^2-4x+6y-12=0$ is $(-23,-28)$.