8,9

A trapezoid with lateral sides $a$ and $b$ is circumscribed around a circle. Find the sum of the squares of the distances from the center of the circle to the vertices of the trapezoid.
Let the lateral sides $AB$ and $CD$ of trapezoid $ABCD$ be equal to $a$ and $b$ respectively, and $O$ be the center of the circle. The center of the circle inscribed in an angle lies on the bisector of this angle, so

$$
\angle BAO + \angle ABO = \frac{1}{2} \angle BAD + \frac{1}{2} \angle ABC = \frac{1}{2} (\angle BAD + \angle ABC) = \frac{1}{2} \cdot 180^{\circ} = 90^{\circ},
$$

which means, $\angle AOB = 90^{\circ}$. From the right triangle $AOB$, we find that

$$
OA^2 + OB^2 = AB^2 = a^2
$$

Similarly, $OC^2 + OD^2 = b^2$. Therefore,

$$
OA^2 + OB^2 + OC^2 + OD^2 = a^2 + b^2.
$$

## Answer

$a^2 + b^2$.