A geometry problem by Galen Buhain

Let $O$ be the center of the radius $r$ circle, $M$ the midpoint of $AB$ and $N$ the midpoint of $CD.$

At this stage we're not sure if the two chords lie on the same side of $O,$ so for now we will assume that they lie on opposite sides of $O$ and we will let $x = |ON|.$ If $x$ comes out to be negative then this will indicate that the two chords lie on the same side of $O.$

Now $\Delta AMO$ is a right triangle with side lengths $|AM| = 5, |MO| = |MN| - |ON| = 17 - x$ and hypotenuse length $|OA| = r.$ By Pythagoras we then have that $r^{2} = (17 - x)^{2} + 5^{2} = 314 - 34x + x^{2}.$

Next, $\Delta CNO$ is a right triangle with side lengths $|CN| = 12, |NO| = x$ and hypotenuse length $|OC| = r.$ By Pythagoras we then have that $r^{2} = x^{2} + 12^{2} = x^{2} + 144.$

Equating these two results for $r^{2}$ gives us that

$314 - 34x + x^{2} = x^{2} + 144 \Longrightarrow 34x = 314 - 144 = 170 \Longrightarrow x = 5.$

(The positive value for $x$ indicates that the two chords are in fact on opposite sides of $O.$ ) Then with this value for $x$ we have that

$r^{2} = x^{2} + 144 = 25 + 144 = 169 \Longrightarrow r = \boxed{13}$ cm.