Two circles with radii $R$ and $r (R > r)$ touch each other internally at A. $Δ A B C$ is an equilateral triangle with $B$ on one circle and $C$ on the other. Then the length of the side of the equilateral triangle $A B C$ is

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$\frac{R r \sqrt{3}}{\sqrt{R^{2} - R r + r^{2}}}$

$\frac{2 R r \sqrt{3}}{\sqrt{R^{2} - R r + r^{2}}}$

$\frac{R r \sqrt{3}}{\sqrt{R^{2} + R r + r^{2}}}$

$\frac{2 R r \sqrt{3}}{\sqrt{R^{2} + R r + r^{2}}}$

answer is B.

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Detailed Solution

$∠ A C Q = ∠ A D P = 90 °$
Hence $Δ A C Q$ and $Δ A D P$ are similar triangles. From $Δ B D A, \frac{B D}{\sin 60 °} = 2 r$
Therefore $B D = \sqrt{3 r}$
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Also, from the similarity of triangles $\frac{A D}{A C} = \frac{A P}{A Q} = \frac{r}{R}$
Therefore $A D = \frac{r x}{R} (∵ A C = x)$
Using cosine rule for $Δ A B D .$
$3 r^{2} = {(B D)}^{2} = {(A B)}^{2} - 2 (A D) (A B) \cos 60 ° + A D^{2}$
$x = \frac{\sqrt{3} R r}{\sqrt{R^{2} - R r + r^{2}}}$