Problem:
Circles C1β and C2β intersect at two points, one of which is (9,6), and the product of their radii is 68. The x-axis and the line y=mx, where m>0, are tangent to both circles. It is given that m can be written in the form abβ/c, where a,b, and c are positive integers, b is not divisible by the square of any prime, and a and c are relatively prime. Find a+b+c.
Solution:
Let r1β and r2β be the radii and A1β and A2β be the centers of C1β and C2β, respectively, and let P=(u,v) belong to both circles. Because the circles have common external tangents that meet at the origin O, it follows that the first-quadrant angle formed by the lines y=0 and y=mx is bisected by the ray through O,A1β, and A2β. Therefore, A1β=(x1β,kx1β) and A2β=(x2β,kx2β), where k is the tangent of the angle formed by the positive x-axis and the ray OA1β. Notice that r1β=kx1β and r2β=kx2β. It follows from the identity tan2Ξ±=1βtan2Ξ±2tanΞ±β that m=1βk22kβ. Now (PA1β)2=(kx1β)2, or
(uβx1β)2+(vβkx1β)2(x1β)2β2(u+kv)x1β+u2+v2β=k2x12β, so =0β
In similar fashion, it follows that
(x2β)2β2(u+kv)x2β+u2+v2=0
Thus x1β and x2β are the roots of the equation
x2β2(u+kv)x+u2+v2=0
which implies that x1βx2β=u2+v2, and that r1βr2β=k2x1βx2β=k2(u2+v2). Thus