I've been reading a bit about inversive geometry, particularly circle inversion. The following is a problem from Hartshorne's classical geometry, which I've been struggling with on and off for a few days.
I figured it would be helpful to show that $TU\perp OA$ first. At best, I tried to label angles according to which ones are congruent with each other. I know $\angle RPS$ and $\angle RQS$ are both right, as they subtend the diameter, so $\angle TPS=\angle UQS$ are both right as well. So $PTUQ$ is a cyclic quadrilateral, and thus $\angle RTQ=\angle PUQ$. Labeling $\angle SPQ$ as $3$ and $\angle PQS$ as $4$, I see that $1+2+3+4$ sum to two right angles.
That's about as far as my observations got me. My hunch is that $RTU$ is an isosceles triangle, and $PU$ is like a line of symmetry, but I'm not sure how to show it, and how to eventually conclude $TU$ meets $OA$ at $A'$, that is, $OA\cdot OA'=r^2$, where $r$ is the radius if $\Gamma$. Thanks for any ideas on how to solve this.