[ Touching circles ] [Common tangent to two circles]

Two touching circles are inscribed in an angle of size $2 \alpha$. Find the ratio of the radius of the smaller circle to the radius of a third circle that touches the first two circles and one of the sides of the angle.
The length of the segment of the common external tangent to two touching circles of radii $r$ and $R$, enclosed between the points of tangency, is $2 \sqrt{r R}$.

## Solution

Let $r$ and $R$ be the radii of the given circles ($r < R$), $O$ and $Q$ their centers, $A$ and $B$ the points of tangency on one side of the angle, $x$ the radius of the third circle, and $C$ its point of tangency on the same side of the angle.

Since

$$
A B = A C + B C, \quad A B = 2 \sqrt{r R}, \quad A C = 2 \sqrt{r x}, \quad B C = 2 \sqrt{R x}
$$

we have

$$
2 \sqrt{r x} + 2 \sqrt{R x} = 2 \sqrt{r R}
$$

From this equation, we find that

$$
x = \frac{r R}{(\sqrt{r} + \sqrt{R})^2}
$$

Then

$$
\frac{r}{x} = \frac{(\sqrt{r} + \sqrt{R})^2}{R} = \left(\frac{\sqrt{r} + \sqrt{R}}{\sqrt{R}}\right)^2 = \left(\sqrt{\frac{r}{R}} + 1\right)^2
$$

Dropping a perpendicular $OP$ from point $O$ to the radius $QB$, in the right triangle $OPQ$

$$
O Q = r + R, \quad P Q = R - r, \quad \angle P O Q = \alpha
$$

Therefore,

$$
R - r = (R + r) \sin \alpha \text{, or } 1 - \frac{r}{R} = \left(1 + \frac{r}{R}\right) \cdot \sin \alpha
$$

From this, we find that

$$
\frac{r}{R} = \frac{1 - \sin \alpha}{1 + \sin \alpha} = \frac{\cos^2 \alpha}{(1 + \sin \alpha)^2}
$$

Thus,

$$
\frac{r}{x} = \left(\frac{\cos \alpha}{1 + \sin \alpha} + 1\right)^2 = \frac{2(1 + \cos \alpha)}{1 + \sin \alpha}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-51.jpg?height=440&width=763&top_left_y=177&top_left_x=636)

## Answer

$\frac{2(1 + \cos \alpha)}{1 + \sin \alpha}$.