Problem:
Circles Ο1β and Ο2β with radii 961 and 625, respectively, intersect at distinct points A and B. A third circle Ο is externally tangent to both Ο1β and Ο2β. Suppose line AB intersects Ο at two points P and Q such that the measure of minor arc PQβ is 120β. Find the distance between the centers of Ο1β and Ο2β.
Solution:
Let R1β=961 and R2β=625 be the radii of Ο1β and Ο2β, respectively, r be the radius of Ο, and β be the distance from the center O of Ο to the line AB. Let O1β and O2β be the centers of Ο1β and Ο2β, respectively. Let X be the projection of O onto line O1βO2β, and let Y be the intersection of AB with line O1βO2β.\
Let the distance between O1β and O2β be d. Then d=O1βYβO2βY. Because AB is a chord in both Ο1β and Ο2β, the power of point Y is the same with respect to both circles. Thus
Note: In the figure shown, it is assumed that the points X,Y,O2β, and O1β occur in that order along the line containing the centers of Ο1β and Ο2β. If the order were different, the same argument with appropriate sign changes would yield the same answer.
OR
Let O,O1β,O2β,R1β,R2β, and r be as in the first solution. Let line OP intersect line O1βO2β at T, and let u=TO2β, v=TO1β and x=PT. Because lines PQ and O1βO2β are perpendicular, lines OT and O1βO2β meet at a 60β angle.
