Question

The polar of $P$ with respect to the circle $x^2+y^2=a^2$ touches the circle $(x-\alpha {)}^2+(y-\beta {)}^2=b^2$; prove that its locus is the curve given by the equation $(\alpha x+\beta y-a^2{)}^2=b^2(x^2+y^2)$.

Answer 1

Let the point $P$ be $(h,k)$

Equation of polar of $P$ w.r.t to $x^2+y^2=a^2$ is $T=0$

$\Rightarrow hx+ky=a^2\Rightarrow hx+ky-a^2=\mathrm{0.....}\left(i\right)$

Now $\left(i\right)$ is tangent to ${(x-\alpha )}^2+{(y-\beta )}^2=b^2$

So perpendicular distance from centre of circle to $\left(i\right)$ is equal to its radius

$\Rightarrow \frac{|h\alpha +k\beta -a^2|}{\sqrt{h^2+k^2}}=b$

Squaring both sides

$\Rightarrow \frac{{(h\alpha +k\beta -a^2)}^2}{h^2+k^2}=b^2\Rightarrow {(h\alpha +k\beta -a^2)}^2=b^2(h^2+k^2)$

Generalising the equation

$\Rightarrow {(\alpha x+\beta y-a^2)}^2=b^2(x^2+y^2)$

Hence proved