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Pair of Tangents from a Point

There are thr...

Question

There are three circles $C_{1}, C_{2}, C_{3}$ with radii $r_{1}, r_{2}, r_{3} (r_{1} < r_{2} < r_{3})$ respectively touching each other externally as shown in the figure and $L_{1}, L_{2}$ are two common tangents to these circles from point $P$ .

Which of the following is(are) correct?

A

r_{2} = \frac{r_{1}^{2} + r_{3}^{2}}{r_{1} + r_{3}}

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B

r_{2}^{2} = r_{1} r_{3}

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C

D E = 2 \sqrt{r_{1} r_{2}}

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D

D E = \sqrt{r_{2}^{2} - r_{1}^{2}}

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Solution

The correct option is C $D E = 2 \sqrt{r_{1} r_{2}}$

$D E = A K = \sqrt{{(r_{1} + r_{2})}_{1}^{2} - {(r_{1} - r_{2})}_{1}^{2}}$
$= 2 \sqrt{(r_{1} r_{2})}$
Let $P D = x$ and $∠ A P D = θ$
$∴ tan θ = \frac{A D}{P D} = \frac{B E}{P E} = \frac{C F}{P F}$

$\Rightarrow \frac{r_{1}}{x} = \frac{r_{2}}{x + 2 \sqrt{(r_{1} r_{2})}} = \frac{r_{3}}{x + 2 \sqrt{(r_{1} r_{2})} + 2 \sqrt{(r_{2} r_{3})}}$

$\Rightarrow \frac{r_{2} - r_{1}}{2 \sqrt{(r_{1} r_{2})}} = \frac{r_{3} - r_{2}}{2 \sqrt{(r_{2} r_{3})}}$

$\Rightarrow r_{2} = \sqrt{(r_{1} r_{3})}$

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