Task B-2.2. Two circles touch each other externally. Their external common tangents form a right angle. If the radius of the smaller circle is $5 \mathrm{~cm}$, calculate the radius of the larger circle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8c954f607b7f1570b01bg-06.jpg?height=607&width=596&top_left_y=1941&top_left_x=732)

Sketch with perpendicular radii drawn.

(2 points)

Let $r$ and $R$ be the radii of the smaller and larger circles, respectively, $A$ be the intersection of the tangents, and $S$ be the center of the larger circle. Since the tangents form a right angle, consider squares where one vertex is point $A$ and the opposite vertex is the center of one of the given circles. Observe the length $\overline{A S}$. On one hand, its length is

$$
|\overline{A S}|=r \sqrt{2}+r+R
$$

On the other hand, $\overline{A S}$ is the diagonal of the larger square in the image, so $|\overline{A S}|=R \sqrt{2}$. It follows that $r \sqrt{2}+r+R=R \sqrt{2}$, or

$$
R=r \frac{\sqrt{2}+1}{\sqrt{2}-1}=5(\sqrt{2}+1)^{2}=15+10 \sqrt{2} \text{ cm}
$$