Question

Let $S_1$ and $S_2$ be two unit circles with centres at $C_1(0,0)$ and $C_2(1,0)$ respectively. Let $S_3$ be another circle of unit radius, passing through $C_1$ and $C_2$ and its centre is above the $x$-axis. If equation of common tangent to $S_1$ and $S_3$, which does not pass through $S_2$, is $ax+by+2=0$, then the value of $a^2-b$ is

Answer 1

Let equation of $S_3$ be
$x^2+y^2+2gx+2fy+c=0$
$S_3$ passes through $C_1(0,0).$
$\Rightarrow c=0$
Also, $S_3$ passes through $C_2(1,0).$
$\Rightarrow g=\frac{-1}{2}$

Given, radius of $S_3$ is $1.$
$\Rightarrow \frac{1}{4}+f^2+0=1$
$\Rightarrow f=\pm \frac{\sqrt{3}}{2}$
But centre is above $x$-axis.
So, $f=-\frac{\sqrt{3}}{2}$
So equation of $S_3$ is ${(x-\frac{1}{2})}^2+{(y-\frac{\sqrt{3}}{2})}^2=1$


Equation of tangent to $S_1$ is
$y=mx\pm \sqrt{1+m^2}\cdots \left(1\right)$
Equation of tangent to $S_3$ is
$y-\frac{\sqrt{3}}{2}=m(x-\frac{1}{2})\pm \sqrt{1+m^2}$
$\Rightarrow y=mx+(\frac{\sqrt{3}}{2}-\frac{m}{2})\pm \sqrt{1+m^2}\cdots \left(2\right)$
$\left(1\right)$ and $\left(2\right)$ represent the same tangent.
$\Rightarrow \frac{\sqrt{3}}{2}-\frac{m}{2}=0$
$\Rightarrow m=\sqrt{3}$

$\therefore$ Required common tangent is
$\sqrt{3}x-y+2=0$
$\therefore a=\sqrt{3},b=-1$
$\Rightarrow a^2-b=4$