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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.

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  • $\begingroup$ Suppose the lines make angles $\alpha$, $\beta$, $\gamma$ (with $\alpha + \beta + \gamma = 2\pi$). Express the radii in terms of trig functions of $\alpha/2$, etc, (and the distance $4$), and simplify. $\endgroup$ – Blue Sep 15 '15 at 20:01
  • $\begingroup$ @Blue I'm not being able to simplify.Please write out atleast a few steps.. $\endgroup$ – user220382 Sep 15 '15 at 21:37
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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,

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

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