Given two circles with radii $R$ and $r(R>r)$, having internal tangency. Find the radius of the third circle that is tangent to the first two circles and their common diameter.

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Let $O$ be the center of a circle with radius $R$, $O_1$ be the center of a circle with radius $r$, $A$ be their point of tangency, $AB$ be their common diameter, $O_2$ be the center of the circle with the unknown radius $x$, $C$ be the point of tangency of the circles with centers $O$ and $O_2$, $D$ be the point of tangency of the circles with centers $O_1$ and $O_2$, and $M$ be the point of tangency with the diameter $AB$ of the circle with center $O_2$.

The line of centers of two tangent circles passes through their point of tangency, so

$$
O O_1 = O A - O_1 A = R - r, \quad O O_2 = O C - O_2 C = R - x, \quad O_1 O_2 = O_1 D + O_2 D = r + x.
$$

Express the area of triangle $O O_1 O_2$ using Heron's formula. Let $p$ be the semi-perimeter of the triangle, then

$$
p = \frac{(R - r) + (R - x) + (r + x)}{2} = R, \quad S_{\triangle O O_1 O_2} = \sqrt{R r x (R - x - r)}
$$

On the other hand,

$$
S_{\triangle O O_1 O_2} = \frac{1}{2} O O_1 \cdot O_2 M = \frac{1}{2} (R - r) x
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-15.jpg?height=106&width=1218&top_left_y=810&top_left_x=426)

## Answer

$\frac{4 r R (R - r)}{(R + r)^2}$.

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