Let A, B, C be points on circle O such that AB is a diameter, and CO is perpendicular to AB. Let P be a point on OA, and let line CP intersect the circle again at Q. If OP = 20 and PQ = 8, find r^2, where r is the radius of the circle.
\(8\cdot \sqrt{20^2+r^2}=\left(r-20\right)\left(r+20\right)\)
\(64\cdot 400+49r^2=r^4-800r^2+400^2=0\)
\(r^2= \left(\sqrt{\frac{849+\sqrt{183201}}{2}}\right)^2\)
\(r^2= approx.638.5, r = approx.25.3\)
