Three circles touch one another externally.The tangents at their points of contact meet a point whose distance from a point of contact is 4.How to find ratio of product of the radii to the sum of radii of the circle?
I assumed the general 2nd degree equations of the three circles as $S_1,S_2$ and $S_3$.Found the three common tangent equations.How to proceed from there?Thanks.
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,