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Q.
Each of the two orthogonal circles $C_{1}$ and $C_{2}$ passes through both the points $(2,0)$ and $(-2,0) .$ If $y=m x+c$ is a common tangent to these circles, then
TS EAMCET 2019
Solution:
Let the equation of the circle be $x^{2}+y^{2}+2 g x+2 f y+c=0$
Since, this circle is passes through $(2,0)$ and $(-2,0)$
$\therefore 4+4 g+c=0\,\,\,...(i)$
$4-4 g+c=0\,\,\,...(ii)$
On solving Eqs. (i) and (ii), we get $g=0, c=-4$
$\therefore $ Equation of circle $x^{2}+y^{2}+2 f y-4=0$
Now, $y=m x+c$ is tangent of circle
$\therefore \left|\frac{-f+c}{\sqrt{1+m^{2}}}\right|=\sqrt{f^{2}+4}$
$\Rightarrow (c-f)^{2}=\left(1+m^{2}\right)\left(f^{2}+4\right)$
$\Rightarrow m^{2} f^{2}+2 c f+4\left(1+m^{2}\right)-c^{2}=0$
Let $f_{1} \,f_{2}$ are roots of equation
$\therefore f_{1}\, f_{2}=\frac{4\left( l +m^{2}\right)-c^{2}}{m^{2}}$
Now, equation of this circle are
$x^{2}+y^{2}+2 f_{1} y-4=0$ and $x^{2}+y^{2}+2 f_{2} y-4=0$
Since, these are orthogonals.
$\therefore 2 f_{1} \,f_{2}=c_{1}+c_{2}$
$\Rightarrow 2 f_{1}\, f_{2}=-8$
$\Rightarrow f_{1}\, f_{2}=-4$
$\Rightarrow \frac{4\left(1+m^{2}\right)-c^{2}}{m^{2}}=-4$
$\Rightarrow c^{2}=4\left(1+2 m^{2}\right)$