Three circles with different radii touch one another externally. The tangents at their points of contact meet at a point whose distance from a point of contact is $4$ where ratio of the product of the radii to the sum of the radii of circles is $λ^{2} : 1$ . Then the value of $| λ |$ is

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Solution

Let $r_{1}, r_{2}$ and $r_{3}$ be the radii of the three circles with centres at $C_{1}, C_{2}$ and $C_{3}$ respectively.
Let the circles touch at $P, Q$ and $R .$

Also, $C_{1} C_{2} = r_{1} + r_{2}, C_{2} C_{3} = r_{2} + r_{3}, C_{3} C_{1} = r_{3} + r_{1}$
Let $O$ be the point whose whose distance from the points of contact is $4$ .
Then, $O$ is the incentre of $Δ C_{1} C_{2} C_{3}$ with $O P = O Q = O R = r$ ,
$r$ being the radius of the incircle.
Hence, $4 = \frac{Δ C_{1} C_{2} C_{3}}{\frac{1}{2} [C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}]} = \frac{S}{s} \dots (1)$
where $s = r_{1} + r_{2} + r_{3} .$
$S^{2} = s (s - C_{1} C_{2}) (s - C_{2} C_{3}) (s - C_{3} C_{1}) = s (r_{1}) (r_{2}) (r_{3})$
Eq. $(1)$ gives
$16 = \frac{S^{2}}{s^{2}} = \frac{s (r_{3}) (r_{1}) (r_{2})}{s^{2}} = \frac{r_{1} r_{2} r_{3}}{r_{1} + r_{2} + r_{3}}$
Hence, the ratio of the product of the radii of the sum of the radii $= 16 : 1 = λ^{2}$
$\Rightarrow | λ | = 4$

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