Let $CP\cap{AB}=\{X\}$, then $AX=XP$ $\Longleftrightarrow$ $\angle{BRA}=\angle{AQB}$. Consequently, $AQRB$ is cyclic $\Longleftrightarrow$ $\angle{BPR}=\angle{PRB}=\angle{PQR}$.

i think that's wrong... or, at least, it's not obvious and there's a considerable number of steps missing. explain.

Dear pleurestique There is nothing wrong. 1. $x=\angle{AQP}=\angle{PAB}=\angle{APX}=\angle{CPR}$. 2. $y=\angle{PBA}=\angle{BQP}=\angle{BCP}$. 3. $\angle{AQB}=x+y=\angle{ARB}=\angle{RPB}$, thus $AQRB$ is cyclic. 4. It remains to see that $\angle{RQB}=x$. I hope this time I presented the data clearly enough for you.

I suppose I'll flesh out Sailor's solution, as it took me a bit to work through it and see why it was true. It's a very slick solution, however. 1. We have $ \angle AQP = \angle PAB$ since both are half the measure of minor arc $ AP$ (since $ AB$ is tangent to $ \Gamma_1$). Then since $ APX$ is isosceles with $ AX = PX$, we have $ \angle PAB = \angle PAX = \angle APX$. Finally, since $ A, P, R$ are collinear and $ X, P, C$ are collinear, we have $ \angle APX = \angle CPR$. 2. We have $ \angle PBA = \angle BQP$ since both are half the measure of minor arc $ BP$. Then $ \angle BQP = \angle BCP$ since $ BPQC$ is cyclic by definition. 3. Then $ \angle AQP + \angle PQB = \angle AQB = x+y$. But also $ \angle ARB = 180^{\circ}-\angle CRP = \angle CPR + \angle PCB = x+y$. Similarly, $ \angle RPB = \angle RPB = 180^{\circ}-\angle BPA = \angle PBA + \angle PAB = x+y$. Since $ \angle AQB = \angle ARB$, $ AQRB$ is cyclic. 4. Since $ AQRB$ is cyclic, $ \angle RQB = \angle RAB = \angle PAB = x$. Then we have $ \angle PQR = \angle RQB + \angle BQP = x+y$. 5. (to finish) Let $ O$ be the circumcenter of $ PQR$. Then $ \angle POR = 2(x+y)$. We have also $ \angle PBR = 180^{\circ}-(\angle RPB + \angle PRB$). But $ \angle PRB = \angle ARB = x+y$ and $ \angle RPB = x+y$ as above, hence $ \angle PBR = 180^{\circ}-2(x+y)$. Thus $ PBRO$ is cyclic. Since $ POR$ is isosceles, we have $ \angle OPR = \angle ORP$. But also $ \angle BPR = \angle BRP$, hence $ \angle BPO = \angle BRO$. But $ \angle BPO + \angle BRO = 180^{\circ}$ since $ PBRO$ is cyclic, hence $ \angle BPO = \angle BRO = 90^{\circ}$. It follows that $ BP$ and $ BR$ are tangent to the circumcircle of $ PQR$.

I think my solution is quicker and it is purely angle chasing.

Edit: I noticed that $Q$ was the center of a spiral similarity that took $APQ$ to $BRQ$. If anyone has a nice solution using spiral similarity, could they post it.

Join AQ and BQ. RBQ = QPC = PAQ (alternate segment theorem). So, ABRQ is cyclic. Again, BQR = BAR = AQP. So, AQB = PQR or, ARB = PQR. Performing similar angle chasing, we get BPR = PQR. This completes the proof.

Assume $| \Gamma_1 | > | \Gamma_2 |$, $$\angle CPQ=\angle PAQ=180^{\circ}-\angle AQR-\angle ARQ=180^{\circ}-\angle PQR-\angle AQP-\angle ARQ=\angle QPR-\angle PAB$$$$\angle CPR=\angle PAB \implies \Delta APB \sim \Delta PRC \implies \Delta BPR \text{ is isosceles}$$$$\angle QAR=\angle QPC=\angle QBR \implies AQRB \text{ is cyclic} \implies \angle BQR=\angle BAR=\angle AQP \implies \angle PQR=\angle AQB=\angle PRB $$

Pretty easy. Three steps: 1) Show that $BP=BR.$ (Motivation: This is true by problem statement and probably will make life a lot easier.) 2) Show that $ABRQ$ is cyclic. (Motivation: We have no idea what the heck $R$ is, and messing around with angles long enough/having a good diagram will make this feel correct.) 3) Reflect $P$ about the midpoint of $AB$ to finish. (Motivation: Cyclic quadrilaterals are cool, $P'$ lies on $(ABRQ),$ and $\triangle ABP'$ is really easy to handle. Okay, full solution. Note that \[\angle BPR=\angle BAP+\angle ABP=\angle AQP+\angle PBQ=\angle AQB\]and that \[\angle BRP=\angle RPC+\angle RCP=180^{\circ}-\angle APC+\angle BCP=\angle AQP+\angle BQP=\angle AQB,\]so $\angle BPR=\angle BRP.$ Now note \[\angle AQB=\angle BAP+\angle ABP=180^{\circ}-\angle APB=\angle BPR=\angle BRP,\]so $ABRQ$ is cyclic. Now reflect $P$ about the midpoint of $AB$ to get $P'.$ Then note \[\angle PQR=\angle P'QR=\angle P'AR=\angle P'AB+\angle BAR=\angle ABR+\angle BAP=\angle BPR.\]

shobber wrote: Let $\Gamma_1$ and $\Gamma_2$ be two circles intersecting at $P$ and $Q$. The common tangent, closer to $P$, of $\Gamma_1$ and $\Gamma_2$ touches $\Gamma_1$ at $A$ and $\Gamma_2$ at $B$. The tangent of $\Gamma_1$ at $P$ meets $\Gamma_2$ at $C$, which is different from $P$, and the extension of $AP$ meets $BC$ at $R$. Prove that the circumcircle of triangle $PQR$ is tangent to $BP$ and $BR$. Could you send an appropriate visual for the problem if I ask?

a kind of different angle-chasing solution

Good practice for angle-chasing. In the diagram attached, note that $\angle ABP=\angle BCP=\angle MQB,$ the first congruence coming from the fact that $AB$ is tangent to $\Gamma_2$ and the second coming from the fact that $P,B,C,Q$ are concyclic(these are the red angles in the diagram). Next, note that $\angle QBC=\angle QPC=\angle QAP,$ the first coming from concyclic and the second coming from the fact that $PC$ is tangent to $\Gamma_1$. From this, it is easy to see that $A,B,R,Q$ are concylic. Notice that we are trying to prove that $\angle BRA=\angle BPR=\angle PQR.$ Let's first try proving that $\angle BRA=\angle PQR,$ and then prove that $\angle BRA=\angle BPR$(equivalent to proving that $BR=BP$). Notice that $\angle BRA=\angle AQB,$ so the first statement is equivalent to proving that $QP$ and $QB$ are isogonal with respect to $Q$. This is easy to see because $\angle AQM=\angle BAR$ from $AB$ being tangent to $\Gamma_1$, and then $\angle BAR=\angle BQR$. Now note that $\angle BQR=\angle BAR$ and $\angle ABP=\angle MQB$(from tangents), so by exterior angle sum, we have that $\angle BPR=\angle BRP$. We are done!!

Claim: $BP = PT$. Proof : $\angle BTP = \angle BEP + \angle EPT = \angle ABP +\angle BAP = \angle BPT$. Claim: $ABTQ$ is cyclic. Proof : $\angle BQA = \angle BQP + \angle PQA = \angle PAB + \angle PBA = \angle BPT = \angle BTP = \angle BTA$. Now we have $\angle PQT + \angle PQB + \angle BQT = \angle PEB + \angle TAB = \angle BTP$ so $BT$ is tangent to $PQT$. we have $BP = BT$ so $BP$ is tangent too.

Another finish after proving that $ABPR$ is cyclic and $BP=BR$ is to consider the midpoint of $PR$, let it be $X$. Then $M$ is the circumcenter of $(ABX)$ since $BX\perp PR$. Note that $$\angle MXA=\angle MAX=\angle AQP.$$So $A,M,X,Q$ are concyclic. This gives $$\angle MQX=\angle MAX=\angle BQR.$$Hence $BQ$ is $Q$-symmedian of $PQR$. This with the fact that $BP=BQ$ gives the desired result. (using @2above diagram)