Problem:
Circles C1 and C2 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 ab/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 r1 and r2 be the radii and A1 and A2 be the centers of C1 and C2, 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,A1, and A2. Therefore, A1=(x1,kx1) and A2=(x2,kx2), where k is the tangent of the angle formed by the positive x-axis and the ray OA1. Notice that r1=kx1 and r2=kx2. It follows from the identity tan2α=1−tan2α2tanα that m=1−k22k. Now (PA1)2=(kx1)2, or
(u−x1)2+(v−kx1)2(x1)2−2(u+kv)x1+u2+v2=k2x12, so =0
In similar fashion, it follows that
(x2)2−2(u+kv)x2+u2+v2=0
Thus x1 and x2 are the roots of the equation
x2−2(u+kv)x+u2+v2=0
which implies that x1x2=u2+v2, and that r1r2=k2x1x2=k2(u2+v2). Thus
k=u2+v2r1r2
and
m=1−k22k=u2+v2−r1r22r1r2(u2+v2)
When u=9,v=6, and r1r2=68, this gives m=4912221, so a+b+c=282.