Consider a circle and a point $$P$$ inside it. We know that the minimum distance between $$P$$ and the points on the circumference is $$2$$ cm, while the maximum distance is $$8$$ cm. What is the circle radius?

As a first step, we can show that the minimum and maximum distances are reached when the points are on the intersection between the circumference and the diameter passing through $$P$$.

Referring to the Figure in the bottom, note that

$$A’P+OP = r,\tag{1}\label{eq562:uno}$$

where $$r$$ is the circle radius. Furthermore, by triangular inequality on $$\triangle OAP$$,

$$AP + OP \geq r.\tag{2}\label{eq562:due}$$ Comparing \eqref{eq562:uno} e \eqref{eq562:due} yields
$AP \geq A’P.$

Analogously,
$$A^{\prime\prime} P=OP+r\tag{3}\label{eq562:tre}$$

while, again because of triangular inequality,

$$AP\leq OP +r.\tag{4}\label{eq562:quattro}$$

By comparing \eqref{eq562:tre} e \eqref{eq562:quattro} we deduce that
$AP \leq A^{\prime\prime}P.$

Thus $$A’P = 2$$ cm, $$A^{\prime\prime}P = 8$$ cm, and the required radius is

$r = \frac{A’P+A^{\prime\prime}P}{2}=5 \ \mbox{cm}$