Power of three!

Let $A,B,C$ denote centers of circles $C_1,C_2,C_3$ respectively and $D,E,F$ points of tangency. When we create triangle $ABC$ , circle centered at $O$ through points $D,E,F$ is incircle of $\triangle ABC$ with radius $r=4$ . Applying formula for the triangle incircle and Heron's formula for triangle area with $s$ denoting triangle's semi-perimeter, we get: $s=((a+b)+(a+c)+(b+c))/2=a+b+c$ $\triangle ABC=\sqrt{s(s-(a+b))(s-(a+c))(s-(b+c))}=\sqrt{(a+b+c)abc}$ $r=\frac{\triangle ABC}{s}=\frac{\sqrt{(a+b+c)abc}}{a+b+c}=\sqrt{\frac{abc}{a+b+c}}$ $\frac{abc}{a+b+c}=r^2=\boxed{16}$