Question

The condition that the circles which passes through the points $(0,a)$ , $(0,-a)$ and touch the line $y=mx+c$ will cut orthogonally is

• A. $c^2=a^2(1+m^2)$
• B. $c^2=a^2(2+m^2)$
• C. $c^2=a^2(3+m^2)$
• D. $c^2=a^2(4+m^2)$

Answer 1

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

Let the centers be $C_1(-g_1,0)$ & $C_2(-g_2,0)$
$\because$They cut each other orthogonally and pass through $(0,a)$ and $(0,-a)$ their equations are
$S_1=x^2+y^2+2g_{1}x-a^2=0,S_2=x^2+y^2+2g_{2}x-a^2=0$
Then $2g_1{g}_2+2f_1{f}_2=c_1+c_2...$ gives
$2g_1{g}_2+2\left(0\right)\left(0\right)=-a^2-a^2....2g_1{g}_2=-2.a^2$
So that $g_1{g}_2=-a^2...1$
Now if a circle $x^2+y^2+2gx-a^2=0$ with center $c(-g_1,0)$ and $r=\sqrt{g^2+a^2}$ where g is a parameter touches the line $y=mx+c$ i.e., $mx-1y+c=0$ then $CP=r$, gives
$\frac{[m(-g)-\left(0\right)+c]}{\sqrt{m^2+1}}=\sqrt{g^2+a^2}$
$\Rightarrow c-mg=\sqrt{g^2+a^2}\sqrt{m^2+1}$
Squaring on both sides
$c^2-2cmg+m^2{g}^2=(g^2+a^2)m^2+(g^2+a^2)$
$g^2+2cmg+[a^2{m}^2+a^2-c^2]=\mathrm{0....2}$
This is a quadratic equation in g
$g_1{g}_2=\left[\frac{(a^2{m}^2+a^2-c^2)}{1}\right]=a^2$ ($\because$ from $1$, $g_1{g}_2=-a^2$)
$\Rightarrow c^2=2a^2+a^2{m}^2$
$\Rightarrow c^2=a^2(2+m^2)$