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 $4$ . Then the ratio of the product of the radii to the sum of the radii of the circles is

16

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4

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8

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\frac{1}{16}

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Solution

The correct option is C $16$
Let $c_{1}, c_{2} a n d c_{3}$ be the centres of three circles of radii $r_{1}, r_{2} a n d r_{3}$ respectively,
then the lengths of the sides of $Δ C_{1} C_{2} C_{3}$ are $C_{1} C_{3} = r_{1} + r_{3}, C_{2} C_{3} = r_{2} + r_{3} a n d C_{1} C_{2} = r_{1} + r_{2}$
Let $O$ be the point of intersection of common tangents in three circles taken in pairs.
Then, $O P = O Q = O R$ .
Also, $O P, O Q a n d O R$ are perpendicular to the sides $C_{1} C_{2}, C_{1} C_{3} a n d C_{2} C_{3}$ respectively.

Therefore,
$O P = O Q = O R = r$ (in radius of $Δ C_{1} C_{2} C_{3})$
$\Rightarrow r = 4$
$\Rightarrow \frac{Area of Δ C_{1} C_{2} C_{3}}{S e m i - p e r i m e t e r} = 4 [\begin{matrix} ∵ r = \frac{Δ}{S} \end{matrix}]$
But Area of $Δ C_{1} C_{2} C_{3} = \sqrt{(r_{1} + r_{2} + r_{3}) r_{1} r_{2} r_{3}} [∵ S = r_{1} + r_{2} + r_{3}]$
$∴ \frac{Area of Δ C_{1} C_{2} C_{3}}{S e m i - p e r i m e t e r} = 4$
$\Rightarrow \frac{\sqrt{(r_{1} + r_{2} + r_{3}) r_{1} r_{2} r_{3}}}{r_{1} + r_{2} + r_{3}} = 4$
$\Rightarrow \frac{r_{1} r_{2} r_{3}}{r_{1} + r_{2} + r_{3}} = 16$

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