Let $A$ is center of $S_1$,$r_1$ is its radius and $m$ intersect $AX$ at $B$ Let $BX=x$ then $AB=|r_1-x|$ $PB^2=r_1^2-(r_1-x)^2=2r_1x-x^2,PX^2=PB^2+BX^2=2r_1x-x^2+x^2=2r_1x$ Same way $YQ=2r_2x$ and so $\frac{XP}{YQ}=\sqrt{\frac{r_1}{r_2}}$

I'm getting S1 and S2 have equal radii, and XYQP is a rectangle. Is this correct?

Define $d(X,\ell)$ as distance from $X$ to $\ell$. As $XY\parallel PQ$, we have $$XP\cdot \sin{\angle PXY}=d(P,XY)=d(Q,XY)=YQ\cdot \sin{\angle XYQ}.$$Let $R$ be the radius of $S_1$ and $r$ the radius of $S_2$. By Law of Sines, $XP=2R\cdot \sin{\angle PXY}$ and $XP=2R\cdot \sin{\angle XYQ}$. Therefore, $$\frac{XP^2}{YQ^2}=\frac{XP\cdot 2R\cdot \sin{\angle PXY}}{YQ\cdot 2r\cdot \sin{\angle XYQ}}= \frac{R}{r}.$$We are done. [asy][asy] size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,X,Y,P,K,Q,R; O=(0,0);X=dir(90);path w=circle(O,1); path s=circle((6,-0.4),1.4);Y=(6,1);P=dir(300);K=foot(P,O,X);Q=intersectionpoints(s,P--100P-99K)[0];R=(6,-0.4); draw(w);draw(s);draw(X--Y);draw(P--Q); dot("$X$",X,N);dot("$Y$",Y,N);dot("$P$",P,S);dot("$Q$",Q,S);dot(O);dot(R); [/asy][/asy]