Question 565708
{{{drawing(300,270,0,10,0,9,


circle(4,4,3),
circle(7,4,2.3),
locate(6,7,P),
locate(6.5,2,Q),
line(6.1183,6.1243, 8,2),
line(6.1183,6.1243, 1.9,1.9),
locate(8,2,B),
locate(1.9,1.9,A),
line(1.9,1.9,8,2)
)
}}}


The easiest solution is probably to draw the segment connecting the centers of the circles (denote Y,Z), as well as segment PQ:

{{{drawing(300,270,0,10,0,9,


circle(4,4,3),
circle(7,4,2.3),
locate(6,7,P),
locate(6.5,2,Q),
line(6.1183,6.1243, 8,2),
line(6.1183,6.1243, 1.9,1.9),
locate(8,2,B),
locate(1.9,1.9,A),
line(1.9,1.9,8,2),
line(4,4,7.05,4.05),
line(6.1183,6.1243,6.2,2),
locate(4,4,Y),
locate(7.05,4.05,Z),
locate(6.3,3.8,R)
)
}}}


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.