Question

Each of the two orthogonal circles $C_1$ and $C_2$ passes through both the points $(2,0)$ and $(-2,0).$ If $y=mx+c$ is a common tangent to these circles, then

• A. $c^2=4\left(1+2m^2\right)$
• B. $c^2=2\left(1+2m^2\right)$
• C. $c^2=1+m^2$
• D. $c^2{m}^2=4\left(1+m^2\right)$

Answer 1

Answer: (A)

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$
Since, this circle is passes through $(2,0)$ and $(-2,0)$
$\therefore 4+4g+c=0...(i)$
$4-4g+c=0...(ii)$
On solving Eqs. (i) and (ii), we get $g=0,c=-4$
$\therefore$ Equation of circle $x^2+y^2+2fy-4=0$
Now, $y=mx+c$ is tangent of circle
$\therefore \left|\frac{-f+c}{\sqrt{1+m^2}}\right|=\sqrt{f^2+4}$
$\Rightarrow (c-f{)}^2=\left(1+m^2\right)\left(f^2+4\right)$
$\Rightarrow m^2{f}^2+2cf+4\left(1+m^2\right)-c^2=0$
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
$x^2+y^2+2f_{1}y-4=0$ and $x^2+y^2+2f_{2}y-4=0$
Since, these are orthogonals.
$\therefore 2f_1{f}_2=c_1+c_2$
$\Rightarrow 2f_1{f}_2=-8$
$\Rightarrow f_1{f}_2=-4$
$\Rightarrow \frac{4\left(1+m^2\right)-c^2}{m^2}=-4$
$\Rightarrow c^2=4\left(1+2m^2\right)$