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

$\frac{a^2}{r^2}$ is equal to 

Let $(\alpha ,\beta )$ be the pole of the given straight line with respect to the circle $x^2+y^2=a^2$ Then, the equation of the polar is

$\alpha x+\beta y-a^2=0$

It is given that $(\alpha ,\beta )$ lies on the circle $x^2+y^2=a^2$

$\therefore {\alpha }^2+{\beta }^2=9a^2$

Since line in (i) touches the circle $x^2+y^2=r^2$

$\therefore \left|\frac{-a^2}{\sqrt{{\alpha }^2+{\beta }^2}}\right|=r\Rightarrow \frac{a^2}{\sqrt{9a^2}}=r\Rightarrow 9r^2=a^2$