2. The chords $AB$ and $CD$ of a circle intersect at a right angle. The lengths of the chords $AD=60$ and $BC=25$ are known. Calculate the radius of the circle.
2. Let $O$ be the center of the circle, and $E$ a point on the circle such that $AE$ is a diameter. Denote $\measuredangle BAC = \varphi$ and $\measuredangle DCA = \psi$ (note that $\varphi + \psi = 90^{\circ}$, according to the problem statement). Then we have $\measuredangle AOD = 2\psi$ and thus $\measuredangle EAD = \measuredangle OAD = \frac{180^{\circ} - 2\psi}{2} = 90^{\circ} - \psi = \varphi$. Therefore, since the peripheral angles over the chords $BC$ and $ED$ are both equal to $\varphi$, it follows that $ED = BC = 25$. Now, using the Pythagorean theorem, we have $(2r)^2 = AE^2 = AD^2 + ED^2 = 60^2 + 25^2 = 3600 + 625 = 4225$, so we get $r = \frac{\sqrt{4225}}{2} = \frac{65}{2}$.