8,9

Chords equal to 12 and 16 are drawn through the ends of the diameter of a circle, intersecting on the circle.

Find the distances from the center of the circle to these chords.
The diameter perpendicular to a chord bisects it.

## Solution

Let $AB$ be the diameter of the circle, $AM=12$ and $BM=16$ be the given chords. Drop perpendiculars $OP$ and $OQ$ to the chords $AM$ and $BM$ respectively. Since the diameter perpendicular to a chord bisects it, points $P$ and $Q$ are the midpoints of these chords, and since $O$ is the midpoint of $AB$, $OP$ and $OQ$ are the midlines of triangle $AMB$.

Therefore,

$$
OP=\frac{1}{2} BM=8, OQ=\frac{1}{2} AM=6
$$

## Answer

8 and 6.