Question

Find the pole of the line $x+y+2=0$ w.r.t. the circle $x^2+y^2-4x+6y-12=0$.

Answer 1

Let the pole be $(h,k)$

Polar equation is $T=S_1$

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

$\Rightarrow hx+ky-2x-2h+3k+3y-12=0$

$\Rightarrow x(h-2)+y(k+3)+(3k-2h-12)=0$

$\Rightarrow x+y\left(\frac{k+3}{h-2}\right)+\left(\frac{(3k-2h-12)}{(h-2)}\right)=0$

This equation is same as

$x+y+2=0$

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

$\Rightarrow k+3=h-2$

$\Rightarrow h-k=5\to \left(1\right)$

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

$\Rightarrow 3k-2h-12=2h-4$

$\Rightarrow 3k-2h-2h=8$

$\Rightarrow 3k-4h=8$

or $-4h+3k=8\to \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$

$\Rightarrow h-k=5$

Multiplying both sides by $4$

$\Rightarrow 4h-4k=20+(-4h+3k=8)\to -k=28$

$\Rightarrow k=-28$

$h=5+k=-23$

$\therefore$ Pole $\to (-23,-28)$