$\left[\begin{array}{l}\text { [Mutual position of two circles] } \\ \text { Triangle inequality }\end{array}\right]$

One circle is inside the other. Their radii are 28 and 12, and the shortest distance between points on these circles is 10. Find the distance between the centers.
Determine the location of the center of the larger circle relative to the smaller circle.

## Solution

It is known that the shortest distance between points of two such circles is the smallest of the two segments of the line of centers enclosed between the circles. Let $O$ and $O_{1}$ be the centers of the larger and smaller circles, respectively. Draw the diameter of the larger circle so that it passes through the center of the smaller circle. Denote the points of intersection with the circles sequentially as $A, B, C$, and $D$. If $D C=10$, then

$$
A B=56-24-10=22, A O_{1}=12+22=34, O A=28
$$

thus,

$$
O O_{1}=34-28=6 .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-19.jpg?height=297&width=368&top_left_y=1189&top_left_x=845)

## Answer

6.