Question

Three circles touch one another externally. The tangents at their points of contact meet at a point whose distance from a point of contact is $4$. Find the ratio of the product of the radii to the sum of the radii of the circles

Answer 1

Suppose the circles have centres at $C_{1}, C_{2}$ and $C_{3}$ with radius $R_{1}, R_{2}$ and $R_{3}$, respectively. Let the circles touch at $A, B$ and $C$. Let the common tangents at $A, B$ and $C$ meet at $O$. We have, $O A=O B=O C=4$ [given]. Now, the circle with centre at $O$ and passing through $A, B$ and $C$ is the incircle of the triangle $C_{1} C_{2} C_{3}$ (because $O A \perp C_{1} C_{2}$.
Therefore, the inradius of $\Delta C_{1} C_{2} C_{3}$ is $4$ .
and $r=\frac{\Delta}{s} .....$(i)

Now, perimeter of a triangle
$ 2 s=R_{1}+R_{2}+R_{2}+R_{3}+R_{3}+R_{1}$
$\Rightarrow 2 s=2\left(R_{1}+R_{2}+R_{3}\right) $
$\Rightarrow s=R_{1}+R_{2}+R_{3} $
and $ \Delta=\sqrt{s(s-a)(s-b)(s-c)} $
$ =\sqrt{\left(R_{1}+R_{2}+R_{3}\right)\left(R_{3}\right)\left(R_{2}\right)\left(R_{1}\right)} $
From Eq. (i), $ 4=\frac{\sqrt{R_{1} R_{2} R_{3}\left(R_{1}+R_{2}+R_{3}\right)}}{R_{1}+R_{2}+R_{3}} $
$\Rightarrow 16=\frac{R_{1} R_{2} R_{3}\left(R_{1}+R_{2}+R_{3}\right)}{\left(R_{1}+R_{2}+R_{3}\right)^{2}} $
$\Rightarrow 16=\frac{R_{1} R_{2} R_{3}}{R_{1}+R_{2}+R_{3}}$