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The circles having radii $r_{1}$ and $r_{2}$ intersect orthogonally length of their common chord is

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$\frac{\sqrt{r_{1}^{2} + r_{2}^{2}}}{r_{1} r_{2}}$

$\frac{\sqrt{r_{1}^{2} + r_{2}^{2}}}{2 r_{1} r_{2}}$

$\frac{r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}}$

$\frac{2 r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}}$

answer is A.

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Detailed Solution

Let $l e t ∠ A B_{1} B_{2} = θ$

$\Rightarrow A D = r_{1} \sin θ$

And $A D = r_{2} \cos θ$

$\Rightarrow A D^{2} (\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}}) = 1$

$\Rightarrow A D =$ $\frac{r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}}$

Length of common chord is $\frac{2 r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}}$

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