Ratio of radii product to radii sum of three touching circles! Try to see what you can't!

If the angles made by the lines at their common point are $2\alpha$ , $2\beta$ , $2 \gamma$ (where their sum is $2\pi$ , then the three radii are pretty clearly $d \tan \alpha \qquad d\tan \beta \qquad d \tan \gamma$ where $d$ is the distance from the point of intersection to any of the points of tangency. (In the given problem, $d=4$ The product of the radii is trivial; as for the sum ...

Since $\tan\alpha = \tan(\pi-\beta-\gamma) = -\tan(\beta+\gamma) = -\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}$

we have $\tan\alpha + \tan\beta + \tan\gamma = \tan\alpha-\tan\alpha(1-\tan\beta\tan\gamma) = \tan\alpha\tan\beta\tan\gamma$

Therefore, $\frac{\text{product of radii}}{\text{sum of radii}} = \frac{d^3\tan\alpha\tan\beta\tan\gamma}{d\tan\alpha\tan\beta\tan\gamma} = d^2$

Simple objective approach could be to consider all circles have same radius.Hence the triangle formed by joining the centers would be equilateral and given distance =in radius of the triangle.From here it can be solved even without using pen and paper.