Question

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 $m$.

Answer 1

Let m = $\tan 2\theta$

As the circles touch the x-axis. Hence,

the equations of the circles becomes

$(x-h{)}^2+(y-r{)}^2=r^2$...........(i)

And also we know,

$r=h\tan \theta$

and also $x=6,y=4$ in equation (i).

$(6-\frac{r}{\tan \theta }{)}^2+(4-r{)}^2=r^2$

$36+\frac{r^2}{{\tan }^2\theta }-\frac{12r}{\tan \theta }+16=0$

$r^2-12r\tan \theta +52ta{n}^2\theta =0$

this is a quadratic equation where

$r_1\ast r_2=52ta{n}^2\theta$

but atp

$52{\tan }^2\theta =\frac{52}{3}$

${\tan }^2\theta =\frac{1}{3}$

$\tan \theta =\frac{1}{\sqrt{3}}$

$\theta =\frac{\pi }{6}$

$m=\tan \frac{\pi }{3}=\sqrt{3}$