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Angles in Alternate Segments

Three Circles...

Question

Three Circles with centres $C_{1}, C_{2}, C_{3}$ and radii $r_{1}, r_{2}, r_{3}$ where $r_{1} < r_{2} < r_{3}$ are placed as shown in the figure.
Then, $r_{2} =$ _________.

A

\sqrt{r_{3} - r_{1}}

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B

\sqrt{r_{3} + r_{1}}

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C

\sqrt{r_{3} r_{1}}

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D

\sqrt{r_{3}^{2} - r_{1}^{2}}

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Solution

The correct option is C $\sqrt{r_{3} r_{1}}$
$Δ P C_{1} A, Δ P C_{2} B, Δ P C_{3} C$ in one angle is $90^{o}$ & angle P is common, then above triangle will be similar.
Hence for $Δ P C_{1} A$ & $Δ P C_{2} B$
$\frac{P A}{r_{1}} = \frac{P B}{r_{2}}$ .........(i)
$Δ P C_{2} B$ & $Δ P C_{3} C$
$\frac{P B}{r_{2}} = \frac{P C}{r_{3}}$ .........(ii)
we know that if tangents are drawn from an external point P to $3$ -circles
$P A \cdot P C = P B^{2}$ or
$\frac{P A}{P B} = \frac{P B}{P C}$
$\frac{r_{1}}{r_{2}} = \frac{r_{2}}{r_{3}}$
$r_{2} = \sqrt{r_{1} r_{3}}$ .

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C_{2}

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r_{1}, r_{2}, r_{3} (r_{1} < r_{2} < r_{3})

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

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