Question

If two circles which pass through the points (0, a) and (0, -a) cut each other orthogonally and touch the straight line $y=mx+c$, then

• A. $c^2=a^2(1+m^2)$
• B. $c^2=a^2|1+m^2|$
• C. $c^2=a^2(2+m^2)$
• D. $c^2=2a^2(1+m^2)$

Answer 1

The correct option is C $c^2=a^2(2+m^2)$

Equation of a family of circles through $(0,a)$ and $(0,-a)$ is $x^2+y^2+2\lambda ax-a^2=0.$ If two members are for $\lambda ={\lambda }_{1}and\lambda ={\lambda }_2$ then since they intersect orthogonally $2{\lambda }_1{\lambda }_2{a}^2=-2a^2\Rightarrow {\lambda }_1{\lambda }_2=-1$
Since the two circles touch the line $y=mx+c$
$\left[\frac{-\lambda am+c}{\sqrt{1+m^2}}\right]={\lambda }^2{a}^2+a^2\Rightarrow a^2{\lambda }^2+2mca\lambda -c^2+a^2(1+m^2)=0\Rightarrow a^2(1+m^2)-c^2=-a^2\Rightarrow c^2=(2+m^2)a^2$