Question

Prove that the two circles which pass through the points $(0,a)$ and $(0,-a)$ and touch the line $y=mx+c$ will cut orthogonally if $c^2=a^2(2+m^2)$.

Answer 1

Let the equation of the circle be
$x^2+y^2+2gx+2fy+\lambda =0$.
Since it passes through $(0,a)$ and $(0,-a)$.
$\therefore a^2=2fa+\lambda =0$
$a^2-2fa+\lambda =0$.
Subtracting, we get $4fa=0,\therefore f=0$; and $\lambda =-a^2$.
Hence the circle becomes
$x^2+y^2+2gx-a^2=0$.
Centre is $(-g,0)$
and radius $=\sqrt{g^2-\lambda }=\sqrt{g^2+a^2}$
It is given that the circle touches the line
$y=mx+c$ or $mx-y+c=0$.
Hence perpendicular from centre should be equal to radius
$\frac{-mg+c}{\sqrt{(m^2+1)}}=\sqrt{g^2+a^2}$.
Square $(c-mg{)}^2=(g^2+a^2\left)\right(m^2+1)$
or $g^2+2gmc+\left\{a^2\right(1+m^2)-c^2\}=0$.
$\therefore g_1{g}_2=a^2(1+m^2)-c^2.....\left(1\right)$
Above is a quadratic in $g$ and gives us two values of $g$ showing that there will be two circles which satisfy the given conditions. If the two values of $g$ be $g_1$ and $g_2$ then the two circles are
$x^2+y^2+2g_{1}x-a^2=0$
and $x^2+y^2+2g_{2}x-a^2=0$.
These two will intersect orthogonally if
$2g_1{g}_2=2f_1{f}_2=c_1+c_2$
$2\left\{a^2\right(1+m^2)-c^2\}+0=-a^2-a^2=-2a^2$
or $a^2(1+m^2)-c^2=-a^2$
or $a^2(2+m^2)=c^2$
is the required condition.