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The easiest solution is probably to draw the segment connecting the centers of the circles (denote Y,Z), as well as segment PQ:
Since PY = (1/2)PA and PZ = (1/2)PB, triangles YPZ and APB are similar with a 1:2 ratio. Additionally, PR = (1/2)PQ (this can be proven by symmetry). Since R lies on YZ, Q must lie on AB.
Or, another way you can prove it is show that the pairs of triangles PRY/PQA and PRZ/PQB are similar. Then, you may let angle PRY = m, angle PQA = m, it follows that angle PRZ = angle PQB = 180-m. Hence, angles PQA + PQB = 180, so A,Q,B are collinear.