Question

Let $S_1$ and $S_2$ are the unit circles with centres at $C_1(0,0)$ and $C_2(1,0)$ respectively. Let $S_3$ is another circle of unit radius, passes 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 cut $S_2,$ is $ax+by+2=0$ then find the value of $(a^2-b)$.

Answer 1

Equation of $S_3$ is $(x-0)(x-1)+(y-0{)}^2$ $+\lambda y=0$
i.e. $x^2+y^2-x+\lambda y=0$
The circle is of unit radius.
$\Rightarrow \sqrt{\frac{1}{4}+\frac{{\lambda }^2}{4}-0}=1\Rightarrow \lambda =\pm \sqrt{3}$
$\therefore$ Circle $S_3$ lie above $x$ -axis $\Rightarrow \lambda =-\sqrt{3}$
i.e. $S_3=x^2+y^2-x-\sqrt{3}y=0$
its centre $S_3=(\frac{1}{2},\frac{\sqrt{3}}{2})$

Slope of line joining $C_1\&C_3=\sqrt{3}$
$\therefore$ Slope of common tangent is $\sqrt{3}$
i.e. $\sqrt{3}x-y+k=0$ is required common tangent and touches circle $S_1$
$\Rightarrow \left|\frac{k}{\sqrt{3+1}}\right|=1\Rightarrow k=\pm 2$
$\therefore$ For the given figure $k=2$
$\therefore$ Required common tangent is
$\sqrt{3}x-y+2=0$
$\therefore a=\sqrt{3},b=-1$
$\therefore (a^2-b)=3+1=4$