The pole of a straight line with respect to the circle $x^2+y^2=a^2$ lies on the circle$x^2+y^2=9a^2$. If the straight line touches the circle $x^2+y^2=r^2$ then 
 

Given circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\left(3\mathrm{a}\right)}^2$ 

Point on this circle is $\left(3\mathrm{a} \mathrm{cos\theta }, 3\mathrm{a} \mathrm{sin\theta }\right)$ 

whose polar w.r.to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$   is ${\mathrm{S}}_1=0$

$3\mathrm{a} \mathrm{cos\theta } \mathrm{x}+3\mathrm{a} \mathrm{sin\theta } \mathrm{y}={\mathrm{a}}^2$

If it  touches the circle  ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$ then 

$$\begin{gathered} \mathrm{r}=\frac{\left|{\mathrm{a}}^2\right|}{\sqrt{9{\mathrm{a}}^2\left(1\right)}} \\ \mathrm{r}=\frac{{\mathrm{a}}^2}{\sqrt{9{\mathrm{a}}^2}} \\ \Rightarrow 3\mathrm{ar}={\mathrm{a}}^2 \\ \Rightarrow 9{\mathrm{r}}^2={\mathrm{a}}^2 \end{gathered}$$