[ Triangle Inequality ]

The least distance from a given point to points on a circle is $a$, and the greatest is $b$. Find the radius.

#
Let point $M$ lie outside a circle with center $O$ (Fig. 1) and $AB=2R$ be a diameter of this circle lying on the line $OM$ ($B$ between $O$ and $M$). If $X$ is any point on the circle, then

$$
MX \leqslant OX + OM = R + OM = AM.
$$

Therefore, $AM$ is the greatest distance from point $M$ to the points on the circle. Since

$$
MX + R = MX + OX \geqslant OM = OB + BM = R + BM,
$$

we have $MX \geqslant BM$. Therefore, $BM$ is the least distance from point $M$ to the points on the circle. Consequently,

$$
R = \frac{AM - BM}{2} = \frac{\hbar - a}{2}.
$$

If point $M$ is inside the circle (Fig. 2), then similarly we find that

$$
R = \frac{b + a}{2}.
$$

If the point lies on the circle, then $a=0$ and $b=2R$, so $R = \frac{b - a}{2} = \frac{b + a}{2}$.

## Answer

$\frac{b - a}{2}$ or $\frac{b + a}{2}$