If the pole of a straight line with respect to the

circle  $e{x}^2+y^2=a^2$ lies on the circle $x^2+y^2=9a^2$ then the straight line

 touches the circle

Let $\left(x_1,y_1\right)$ be the pole of the given straight line

  with respect to $x^2+y^2=a^2$  Then the equation of the polar i.e.

the straight line is

             $x{x}_1+y{y}_1=a^2$                             …….(i)

 It is given that  $\left(x_1,y_1\right)$ lies on $x^2+y^2=9a^2$ 

       $\therefore x_1^2+y_1^2=9a^2$

Now, length of the perpendicular from the centre  $(0, 0)$  of the

circle $x^2+y^2=\frac{a^2}{9}$ on 9 on (i) is 

$\left|\frac{0x+0y_1-a^2}{\sqrt{x_1^2+y_1^2}}\right|=\frac{a^2}{\sqrt{x_1^2+y_1^2}}=\frac{a^2}{\sqrt{9a^2}}=\frac{a}{3}$                                    [Using (ii)]

Clearly, length of the perpendicular from the centre O $(0,0 )$  of the

circle $x^2+y^2=\frac{a^2}{9}$  to the line (i) is equal to the radius of the

circle 

Hence, the straight line touches the circle $x^2+y^2=\frac{a^2}{9}$