The pole of the line x a + y b = 1 with respect of the circle x 2 + y 2 = c 2 is

Let $P\left(x_1,y_1\right)$ be a pole with respect to the circle $x^2+y^2=c^2$

Then polar equation is given by,

$T=0$

 $\Rightarrow x{x}_1+y{y}_1-c^2=0\to \left(1\right)$

$\Rightarrow \frac{x{x}_1}{c^2}+\frac{y{y}_1}{c^2}-1=0$

Since $P\left(x_1,y_1\right)$ lies on $\frac{x}{a}+\frac{y}{b}=1\Rightarrow \frac{x_1}{a}+\frac{y_1}{b}-1=0\to \left(2\right)$

Now comparing with equation $\left(1\right)$ and equation $\left(2\right)$ we get point $\left(x_1,y_1\right)\equiv \left(\frac{c^2}{a},\frac{c^2}{b}\right)$