Question

Two circles pass through the points $(0,a)$ and $(0,-a)$ and touch the line $y=mx+c$ If they intersect orthogonally, then $\frac{c^2}{a^2}$ is

• A. $m^2$
• B. $m^2+1$
• C. $m^2+2$
• D. $m^2-1$

Answer 1

The correct option is C $m^2+2$

The circles through $(0,\pm a)$ are $x^2+y^2-a^2+\lambda x=0,\lambda \in R$ ..............$\left(1\right)$
Consider the two circles $x^2+y^2+{\lambda }_{1}x-a^2=0,x^2+y^2+{\lambda }_{2}x-a^2=0$
if they intersect orthogonally, $\frac{{\lambda }_1{\lambda }_2}{2}=-2a^2$
or ${\lambda }_1{\lambda }_2=-4a^2$ ................$\left(2\right)$
If $y=mx+c,$ eqn$\left(1\right)$ becomes
$x^2+{(mx+c)}^2+\lambda x-a^2=0$
or $x^2(m^2+1)+(\lambda +2cm)x+c^2-a^2=0$
Discriminant$=0\Rightarrow {(\lambda +2cm)}^2-4(m^2+1)(c^2-a^2)=0$
$\therefore {\lambda }^2+4cm\lambda +4(a^2(m^2+1)-c^2)=0$
If ${\lambda }_1,{\lambda }_2$ are the roots, then
${\lambda }_1{\lambda }_2=4(a^2(m^2+1)-c^2)$ ........$\left(3\right)$
From eqns$\left(2\right)$ and $\left(3\right)$
$\Rightarrow -a^2=a^2(m^2+1)-c^2$
or $-a^2(1+m^2+1)=-c^2$
or $\frac{c^2}{a^2}=m^2+2$