The circles 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 D $\frac{2 r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}}$
Two circles intersect orthogonality means $r_{1}$ through point of contact makes right angle with $r_{2}$ through that point
So the line intersecting has length $\sqrt{r_{1}^{2} + r_{2}^{2}}$
Common chard length $= h$
By equating area or triangle formed in 2 different way
$\frac{1}{2} r_{1} r_{2} = \frac{1}{2} \times \frac{h}{2} (\sqrt{r_{1}^{2} + r_{2}^{2}})$
$\Rightarrow h = \frac{2 r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}}$

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