Question

Prove that the polar of a given point with respect to any one of the circles $x^2+y^2-2kx+c^2=0$, where $k$ is variable, always passes through a fixed point, whatever be the value of $k$.

Answer 1

For $x^2+y^2-2kx+c^2=0$

Equation of polar w.r.t to any point $P(a,b)$ is $T=0$

$ax+by-2k\left(\frac{x+a}{2}\right){+c}^2=0ax+by-kx-ka+c^2=0ax+by+c^2-k(x+a)=0$

This represents a family of line of form $P+\lambda Q=0$

So it always passes through the point of intersection of $ax+by+c^2=\mathrm{0...}\left(i\right)$ and $x+a=\mathrm{0......}\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow x=-a,y=\frac{a^2-c^2}{b}$

So the family of lines always passes through $(-a,\frac{a^2-c^2}{b})$