Problem 744

Prove that the shortest distance between points of two circles, one lying outside the other, is the segment of the line of centers enclosed between the circles.

## Solution

We will prove that the shortest distance between points of two circles, one lying outside the other, is the segment of the line of centers enclosed between the circles.

Let $O_{1}$ and $O_{2}$ be the centers of the circles, and the line of centers intersects the circles at points $A$ and $B$, such that both $A$ and $B$ lie between $O_{1}$ and $O_{2}$. Then, if $X$ and $Y$ are other points on these circles, we have

$$X O_{1} + X Y + Y O_{2} > O_{1} O_{2} = A O_{1} + A B + B O_{2}$$

Therefore, $X Y > A B$.

Let $A M$ and $B N$ be the diameters of the circles, and $X$ and $Y$ be points on the circles different from $M$ and $N$. Then

$$X Y < X O_{1} + O_{1} O_{2} + Y O_{2} = M O_{1} + O_{1} O_{2} + N O_{2} = M N \text{. }$$

In our problem, $A B = 3$ and $M N = 2 + 8 + 3 = 13$.

## Answer

3 and 13.