Please consider this problem:
http://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_14
Now look at solution 2 - it assumes that A, B, and R are co-linear, but does not prove it. I tried to prove it myself but did not succeed. Can anyone help me prove it?
The Problem can be solved without using this fact. By setting P as the origin and QR as the x-axis one can easily get an equation containing PQ only (although it is tricky to solve). The fact that A, B, and R need not be assumed co-linear furthers my suspicion that A, B and R are not necessarily co-linear, if that makes any sense.
If my Mathematica calculations are correct, $A$, $B$, and $R$ are collinear.
The coordinates of $P$ are $(\frac{43}{6},\frac{\sqrt{455}}{6})$. If $\overline{AB}$ is extended to $R=(18,0)$, $A$, $B$, and $R$ are necessarily collinear, and a calculation shows that $\overline{PR} = \sqrt{130}$. The point on Circle $A$ at distance $\sqrt{130}$ from $P$ is $(-\frac{11}{3},\frac{\sqrt{455}}{3})$. It's on the line extending $RP$, so it must be $Q$.