geometryyyyy 4

Let r be the radius.

Then OA = OB = OC = r.

By Pythagorean theorem, 

$CP^2 = OP^2 + OC^2\\ CP^2 = 20^2 + r^2\\ CP = \sqrt{r^2 + 400}$

Then,

$AP = AO - OP = r - 20\\ BP = BO + OP = r + 20$

Consider the power of point P.

$CP \cdot PQ = AP \cdot PB\\ 7\sqrt{r^2 + 400} = (r - 20)(r + 20)\\ 49(r^2 + 400) = (r^2 - 400)^2\\ (r^2)^2 - 800r^2 + 160000 - 49r^2 - 19600 = 0\\ (r^2)^2 - 849r^2 + 140400 = 0$

Using quadratic formula,

$r^2 = \dfrac{849 \pm \sqrt{849^2 - 4(140400)}}{2} \\ r^2 = 225\text{(rejected) or }\boxed{r^2 = 624}$

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 = 7, find r^2, where r is the radius of the circle.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

OP = 20        PQ = 7         CP = sqrt(r² + 20²)        AP = r - 20         PB = r + 20

PQ * CP = AP * PB

7 * sqrt(r² + 20²) = (r - 20)(r + 20)

r = 4√39           r² = 624