If the polar of point on the circles ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ w.r.t ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{b}}^2$ touching the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{c}}^2$ then show that a,b,c are in G.P

Given circles are ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2\dots \left(1\right)$
$\begin{array}{c}{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{b}}^2\dots ..\left(2\right)\\ {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{c}}^2\dots \dots \left(3\right)\end{array}$
All are concentric. Let P(x₁, y₁) be any point on circle (1)
Polar of P(x₁, y₁)wrt circle (2) is S₁= 0
$\Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1-{\mathrm{b}}^2=0 ...\left(4\right)$

It touches circle (3)
$\perp$^er distance from centre c(0,0) to line (4) = radius r of circle (3)
$\Rightarrow \mathrm{d}=\mathrm{r}$
$\Rightarrow \frac{\left|\mathrm{C}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=\mathrm{r}\Rightarrow \frac{\left|-{\mathrm{b}}^2\right|}{\sqrt{{\mathrm{x}}_1^2+{\mathrm{y}}_1^2}}=\mathrm{c}$
$\because \text{ Radius of circle }\left(3\right)\text{ is }\mathrm{c}\mathrm{\&}\text{ centre is }(0,0)$
$\begin{array}{l}\Rightarrow \frac{{\mathrm{b}}^2}{\sqrt{{\mathrm{a}}^2}}=\mathrm{c}\left(\because {\mathrm{x}}_1^2+{\mathrm{y}}_1^2={\mathrm{a}}^2\right)\\ \Rightarrow \frac{{\mathrm{b}}^2}{\mathrm{a}}=\mathrm{c}\Rightarrow {\mathrm{b}}^2=\mathrm{ac}\Rightarrow \mathrm{a},\mathrm{b},\mathrm{c}\text{ are in G.P }\end{array}$