Let  $S_1=0$ and $S_2=0$ be two circles intersecting at $P(6,4)$  and both are tangent to x-axis and line $y=mx(wherem>0)$. If product of radii of the circles $S_1=0$ and $S_2=0$ is $\frac{52}{3}$ , then find the value of  $m^2$.

Let  $m=\tan 2\theta$
    As the circles touch the x-axis. Hence, the equations of the circles becomes 
 ${(x-h)}^2+{(y-r)}^2=r^2\mathrm{......................}\left(i\right)$       
    And also we know,   $r=h\tan \theta$
     And also  $x=6,y=4$ in equation (i).
 $\begin{array}{l}{(6-\frac{r}{\tan \theta })}^2+{(4-r)}^2=r^2\\ 36+\frac{r^2}{{\tan }^2\theta }-\frac{12r}{\tan \theta }+16=0\\ r^2-12r\tan \theta +52{\tan }^2\theta =0\end{array}$
    This is a quadratic equation where
   $r_1\times r_2=52{\tan }^2\theta$
    But atp 
 $52{\tan }^2\theta =\frac{52}{3}$
 $\begin{array}{l}{\tan }^2\theta =\frac{1}{3}\\ \tan \theta \frac{1}{\sqrt{3}}\\ \theta =\frac{\pi }{6}\\ m=\tan \frac{\pi }{3}=\sqrt{3}\\ m^2=3\end{array}$