The answer is $\frac{r_1}{r_2}$. Let the common external tangents of $\omega_1$ and $\omega_2$ meet at $E$ and the common internal tangents of $\omega_1$ and $\omega_2$ meet at $I$. Let $A'$ and $B'$ be the antipodes of $A$ and $B$ on $\omega_1$ and $\omega_2$. Note that $AB'$ and $A'B$ meet at $E$, and that $P$ lies on $A'B$ and $Q$ lies on $AB'$ because $AA'P$ and $BB'Q$ are right triangles. As $\angle AO_1P+\angle BO_2Q=180^\circ$, we have that $\angle AA'P+\angle BB'Q=90^\circ$, so $AA'P$ is similar to $B'BQ$. Hence, $A'AB$ is similar to $ABB'$. As $P$ and $Q$ are the feet from $A$ to $A'B$ and $B$ to $AB'$, $APB$ and $AQB$ are congruent right triangles. As $\left(\frac{AP}{PB}\right)^2=\left(\frac{A'A}{AB}\right)^2=\frac{A'A}{B'B}=\frac{r_1}{r_2}$, $PQ$ passes through $I$. Hence, $P$ and $Q$ are images of each other under the homothety at $I$ swapping $\omega_1$ and $\omega_2$. But this is also true for $X$ and $Y$, from which it follows that $\frac{PX}{QY}=\frac{r_1}{r_2}$.

ABCDE wrote: The answer is $\frac{r_1}{r_2}$. Let the common external tangents of $\omega_1$ and $\omega_2$ meet at $E$ and the common internal tangents of $\omega_1$ and $\omega_2$ meet at $I$. Let $A'$ and $B'$ be the antipodes of $A$ and $B$ on $\omega_1$ and $\omega_2$. Note that $AB'$ and $A'B$ meet at $E$, and that $P$ lies on $A'B$ and $Q$ lies on $AB'$ because $AA'P$ and $BB'Q$ are right triangles. As $\angle AO_1P+\angle BO_2Q=180^\circ$, we have that $\angle AA'P+\angle BB'Q=90^\circ$, so $AA'P$ is similar to $B'BQ$. Hence, $A'AB$ is similar to $ABB'$. As $P$ and $Q$ are the feet from $A$ to $A'B$ and $B$ to $AB'$, $APB$ and $AQB$ are congruent right triangles. As $\left(\frac{AP}{PB}\right)^2=\left(\frac{A'A}{AB}\right)^2=\frac{A'A}{B'B}=\frac{r_1}{r_2}$, $PQ$ passes through $I$. Hence, $P$ and $Q$ are images of each other under the homothety at $I$ swapping $\omega_1$ and $\omega_2$. But this is also true for $X$ and $Y$, from which it follows that $\frac{PX}{QY}=\frac{r_1}{r_2}$. Well, the answer may be $\sqrt{\frac{r_1}{r_2}}$, according to a Chinese Geometry teacher.

I see, let me fix my solution. Everything is the same until the last two sentences. But $P$ and $Q$ are not images of each other under the homothety as $IP=IQ$. Instead, they are images under the inversion at $I$ swapping $\omega_1$ and $\omega_2$. Thus, $PXQY$ is cyclic with $PQ$ and $XY$ meeting at $I$, so $\frac{PX}{QY}=\frac{IX}{IQ}=\frac{IA}{IQ}=\sqrt{\frac{IA}{IB}}=\sqrt{\frac{r_1}{r_2}}$.

Answer is $$ \frac{PX}{QY} = \sqrt{\frac{r_1}{r_2}}$$Let $\Omega$ be the circle with diameter $AB$. Then $O_1A,O_2B$ are tangents to $\Omega$, meaning $\Omega$ is orthogonal to both $\omega_2,\omega_2$. More precisely, $O_1P$ and $O_2Q$ are also tangents to $\Omega$. Let $AB \cap XY \cap O_1O_2 = T$. Look at $\Omega$. We know $AA \cap PP, BB \cap QQ, AB \cap PQ$ should be collinear, implying $T \in PQ$. [asy][asy] size(250); pair A=dir(130),B=-A,P=dir(185),Q=2*foot(P,A,B)-P,O1=2*A*P/(A+P),O2=2*B*Q/(B+Q),T=extension(A,B,O1,O2),X=2*foot(A,O1,O2)-A,Y=2*foot(B,O1,O2)-B; draw(unitcircle,red); dot("$A$",A,dir(90)); dot("$B$",B,dir(-90)); dot("$P$",P,dir(-130)); dot("$Q$",Q,dir(20)); dot("$O_1$",O1,dir(O1)); dot("$O_2$",O2,dir(O2)); dot("$T$",T,dir(90)); dot("$X$",X,dir(-70)); dot("$Y$",Y,dir(Y)); draw(X--Y^^A--B^^O1--O2,purple); draw(P--Q,green); draw(P--X^^Y--Q,brown); draw(A--O1--P^^Q--O2--B,blue); draw(circle(O1,abs(O1-A))); draw(circle(O2,abs(O2-B))); label("$\omega_1$",2.25*O1-1.25*A); label("$\omega_2$",2.1*O2-1.1*Y); label("$\Omega$",1.15*dir(-90)); [/asy][/asy] Now the condition $\angle AO_1P + \angle BO_2Q$ would force $AP = AQ$, and hence $AB$ is the perpendicular bisector of segment $PQ$. Particularly, $TP = TQ$. Now, \begin{align*} \frac{PX}{QY} &= \frac{r_1 \sin \angle PAX}{r_2 \sin \angle QBY} = \frac{r_1}{r_2} \cdot \frac{\sin \angle TXP}{\sin \angle TYQ} \qquad \text{and} \\ \frac{PX}{QY} &= \frac{TP \cdot \frac{\sin \angle PTX}{\sin \angle TXP}}{TQ \cdot \frac{\sin \angle QTY}{\sin \angle TYQ}} = \frac{\sin \angle TYQ}{\sin \angle TXP} \\ \implies \left( \frac{PX}{QY} \right)^2 &= \frac{PX}{QY} \cdot \frac{PX}{QY} = \left( \frac{r_1}{r_2} \cdot \frac{\sin \angle TXP}{\sin \angle TYQ} \right) \cdot \frac{\sin \angle TYQ}{\sin \angle TXP} = \frac{r_1}{r_2} \\ \therefore ~ \frac{PX}{QY} &= \sqrt{\frac{r_1}{r_2}} \end{align*}