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, $OA=OB=OC=4$ [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 $OA\perp C_1{C}_2$.
Therefore, the inradius of $\Delta C_1{C}_2{C}_3$ is $4$ .
and $r=\frac{\Delta }{s}.....$(i)

Now, perimeter of a triangle
$2s=R_1+R_2+R_2+R_3+R_3+R_1$
$\Rightarrow 2s=2\left(R_1+R_2+R_3\right)$
$\Rightarrow s=R_1+R_2+R_3$
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), $4=\frac{\sqrt{R_1{R}_2{R}_3\left(R_1+R_2+R_3\right)}}{R_1+R_2+R_3}$
$\Rightarrow 16=\frac{R_1{R}_2{R}_3\left(R_1+R_2+R_3\right)}{{\left(R_1+R_2+R_3\right)}^2}$
$\Rightarrow 16=\frac{R_1{R}_2{R}_3}{R_1+R_2+R_3}$