The circle having radii $r_{1}$ and $r_{2}$ intersect orthogonally. Length of their common chord is

\frac{2 r_{1} r_{2}}{\sqrt{(r_{1}^{2} + r_{2}^{2})}}

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\frac{\sqrt{(r_{1}^{2} + r_{2}^{2})}}{2 r_{1} r_{2}}

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\frac{r_{1} r_{2}}{\sqrt{(r_{1}^{2} + r_{2}^{2})}}

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\frac{\sqrt{(r_{1}^{2} + r_{2}^{2})}}{r_{1} r_{2}}

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Solution

The correct option is A $\frac{2 r_{1} r_{2}}{\sqrt{(r_{1}^{2} + r_{2}^{2})}}$
Let length of common chord is $h$
Now equating area of triangle $P C_{1} C_{2}$ in two way,
$\Rightarrow \frac{1}{2} C_{1} C_{2} \times \frac{h}{2} = \frac{1}{2} r_{1} r_{2}$
Now $C_{1} C_{2} = \sqrt{r_{1}^{2} + r_{2}^{2}}$ , Since $Δ P C_{1} C_{2}$ is right angled at $P$
$∴ h = \frac{2 r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}}$

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