Problem:
Circles ω1 and ω2 with radii 961 and 625, respectively, intersect at distinct points A and B. A third circle ω is externally tangent to both ω1 and ω2. Suppose line AB intersects ω at two points P and Q such that the measure of minor arc PQ is 120∘. Find the distance between the centers of ω1 and ω2.
Solution:
Let R1=961 and R2=625 be the radii of ω1 and ω2, respectively, r be the radius of ω, and ℓ be the distance from the center O of ω to the line AB. Let O1 and O2 be the centers of ω1 and ω2, respectively. Let X be the projection of O onto line O1O2, and let Y be the intersection of AB with line O1O2.\
Let the distance between O1 and O2 be d. Then d=O1Y−O2Y. Because AB is a chord in both ω1 and ω2, the power of point Y is the same with respect to both circles. Thus
Substituting the first equality into the second one and subtracting yields
2r(R1−R2)=d(O1X+O2X)−d(O1Y+O2Y)=2dXY=2dℓ
which shows that
dR1−R2=rℓ=cos(60∘)=21
Therefore d=2(R1−R2)=2(961−625)=672.
Note: In the figure shown, it is assumed that the points X,Y,O2, and O1 occur in that order along the line containing the centers of ω1 and ω2. If the order were different, the same argument with appropriate sign changes would yield the same answer.
OR
Let O,O1,O2,R1,R2, and r be as in the first solution. Let line OP intersect line O1O2 at T, and let u=TO2, v=TO1 and x=PT. Because lines PQ and O1O2 are perpendicular, lines OT and O1O2 meet at a 60∘ angle.
