- Let $ \omega_1$ and $ \omega_2$ be the given circles and $ w$ is tangent to $ \omega_1$,$ \omega_2$ at $ B,A$,respectively.Note that $ w$ is tangent either internally,or externally to both of $ \omega_1$ and $ \omega_2$,otherwise it can not be intersected by $ XY$.Suppose that $ w$ is externally tangent to $ \omega_1$ and $ \omega_2$. Denote intersection points of $ XY$ with $ w$,as $ C$ and $ D$,such that $ D$ lies between $ l_1$ and $ l_2$ and $ C$ doesn't lie there and distance between $ C$ and $ l_2$ is greater than distance between $ C$ and $ l_1$.Let $ K\in DB\cap\omega_1$,$ N\in DA\cap\omega_2$,$ M\in CA\cap\omega_2$ and $ L\in CB\cap\omega_1$.Now if we prove that $ KN$ and $ LM$ are common tangent lines to $ \omega_1$ and $ \omega_2$,we will be done. -Consider the homothety with center at $ B$,which turns $ w$ to $ \omega_1$.Then obviously $ D\rightarrow K$ and $ C\rightarrow L$,thus $ KL\| CD$.As we know $ CB\cdot CL = CX\cdot CY = CA\cdot CM$,hence quadrilateral $ CBLM$ is cyclic,so $ \widehat{MLB} = \widehat{BAC} = \widehat{LKB}$.From where we conclude that $ LM$ is tangent to $ \omega_1$.In similar way we prove that $ LM$ is also tangent to $ \omega_2$,so $ LM$ and $ KN$ are common tangent lines to $ \omega_1$ and $ \omega_2$ and obviously $ K,L,M,N$ are fixed points.So we are done. -The case when $ w$ is internally tangent to $ \omega_1$ and $ \omega_2$ is totally similar to the previous case.

I want to understand your solution and I don't understand who's $ l_1$ and $ l_2$. Please answer me !

Here's a fix for the typos (I think this solution is the same?) Let CA intersect the circle tangent at A again at E, CB intersect the other circle again at F. Let BA intersect the circle containing E again at B'. EB'A ~ CBA, so mABC=mEB'A. C is on the radical axis of the 2 circles, so EABF is cyclic, and we have mABC=mFEA, and EF is a tangent to the circle containing E. Applying the same argument 4 times over, or extending DA, DB and taking note of the chords being parallel to the radical axis gives us that the given lines intersect the 2 circles at the points of tangency of the common tangents.

Too many circles. Invert! Invert about a circle centered at X. Then the the two circles intersecting at X become lines intersecting the radical axis at Y. Easy to show that AC always makes the same angle with the radical axis since A,B,C,D move uniformly with the circle, so then we can easily construct one of our points P so as to make a trapezoid XPAC, where P lies on what is now the inverted first circle. Then XPAC is cyclic, and is always cyclic, corresponding to the desired condition in the original diagram. Repeat with the other 3 lines/circles in the inverted diagram, and we have the result.

Hm... 90% perspiration (drawing the diagram ), 10% inspiration (oh yeah, the four points obviously are the common tangents' tangency points ) Let $MN$ be a common tangent of the two given circles $\omega_1$ and $\omega_2$, and call $\omega$ the third circle. Homotheties taking $\omega_1$, $\omega_2$ to $\omega$ take $M, N$ to $D_1, D_2$ respectively and $MN$ to a tangent of $\omega$ parallel to $MN$, $l$. Because there is only one such tangent $l$, we have $D_1 = D_2$. If $F$ is a point on $l$ such that $\angle MND = \angle NDF$, then incidentally $\angle D_1 AB = \angle ND_1 F$, so $MNBA$ is cyclic. Thus $D_1$ lies on $XY$ by radical axis theorem, so $D_1 = D$ or $C$. Similar result holds for other common tangent.

No one has posted a diagram yet so I'll post mine(my solution uses the same ideas): [asy][asy] size(10cm); path circ1, circ2, dcirc1, dcirc2, circ, radaxis; real r1=5, r2=4.2, h=2, r=3.8, e=.7; pair O1= (-sqrt(r1^2-h^2),0), O2=(sqrt(r2^2-h^2),0); circ1=circle(O1,r1); circ2=circle(O2,r2); dcirc1=circle(O1,r1+r); dcirc2=circle(O2,r2+r); pair [] x=intersectionpoints(circ1, circ2); radaxis=(0,10^7) -- (0,-10^7); pair [] z=intersectionpoints(dcirc1, dcirc2); circ=circle(z[1],r); pair A=r1*dir(z[1]-O1)+O1, B=r2*dir(z[1]-O2)+O2; pair [] c=intersectionpoints(radaxis, circ); pair H=(r1*O2-r2*O1)/(r1-r2); pair M1=conj((O1-H)^2/(O1+(0,r1)-H))+H, M2=conj((O2-H)^2/(O2+(0,r2)-H))+H; pair N2=(O2-H)^2/(O2+(0,r2)-H)+conj(H), N1=(O1-H)^2/(O1+(0,r1)-H)+conj(H); draw(circ1); draw(circ2); draw(x[0]+(0,e) -- c[1]-(0,e),mediumred); draw(circ,heavyblue); draw(M1--M2,heavygray); draw(N2--N1,heavygray); draw(M1--c[1]--M2,heavygreen); draw(N1--c[0]--N2,heavygreen); draw(circumcircle(A,B,M1),dashed+heavycyan); dot("$A$", A, 1.5*dir(80)); dot("$B$", B, 1.5*dir(110)); dot("$C$", c[0], dir(310)); dot("$D$", c[1], dir(220)); dot("$X$", x[0], dir(165)); dot("$Y$", x[1], dir(190)); dot("$O$", z[1], dir(60)); dot("$M_1$", M1, dir(120)); dot("$N_1$", N1, dir(260)); dot("$M_2$", M2, dir(40)); dot("$N_2$", N2, dir(300)); [/asy][/asy] Call the two original circles $\omega_1$ and $\omega_2$, with $A$ on $\omega_1$ and $B$ on $\omega_2$. Let $M_1$ be the second intersection of $\overline{AD}$ with $\omega_1$, and let $M_2$ be the second intersection of $\overline{BD}$ with $\omega_2$. Define $N_1$ and $N_2$ analogously, with point $D$ replaced with point $C$. Since $A$ is the center of the homothety with negative scale factor sending $\omega$ to $\omega_1$, and this homothety sends $\overline{CD}$ to $\overline{N_1M_1}$, $\angle ACD=\angle AN_1M_1$. Notice that $\overline{XY}$ is the radical axis of $\omega_1$ and $\omega_2$, so $DA\cdot DM_1=DB\cdot DM_2$ and it follows that quadrilateral $ABM_2M_1$ is cyclic. Thus, $$\angle M_2M_1A=\angle ABD=\angle ACD=\angle AN_1M_1,$$so $\overline{M_1M_2}$ is tangent to $\omega_1$. We can repeat this three times to find $\overline{M_1M_2}$ and $\overline{N_2N_2}$ are tangent to $\omega_1$ and $\omega_2$, so points $M_1$, $M_2$, $N_1$, and $N_2$ are fixed. These are the desired four points and we are done. (This is another configuration where the circles are internally tangent, but the same proof still holds.)

Let $\omega_1, \omega_2$ be the names of the two circles with $A$ corresponding to $\omega_1$ and $B$ corresponding to $\omega_2$. Let $E$ and $F$ be the intersection of $BD$ and $BC$ respectively with $\omega_2$. Lastly, let $O_1,O_2,O_3$ be the centers of $\omega_1,\omega_2,$ and the circle tangent to both, respectively. We claim that $E$ and $F$ are fixed, this will solve the problem because we can apply a similar thing to the other two points. Notice that by homothety triangles $CBD$ and $FBE$ are similar, and we get that $CD\parallel FE$. In addition, by 2021 AIME I #13, we get that given the two circles $\omega_1$ and $\omega_2$, the angle $\angle CO_3D$ is fixed, which means that by homothety, the angle $\angle EO_2F$ is fixed. Therefore minor arc $FE$ of $\omega_2$ is fixed. Combined with the parallel condition and the fact that $E$ and $F$ lie on $\omega 2$, this proves that $E$ and $F$ are fixed, and we are done.

Let our circles be $\omega_1$ and $\omega_2$ and their common tangents touch them at point's $S,T$ and $K,M$ respectively. Let common tangents meet at $P$. Step1 : $SKBA$ is cyclic. By Homothety of circles $\omega_1$ and $\omega_2$ we have $SK$ and $AB$ both meet at $P$ and $\angle PBK = \angle PSA$. Step2 : $C,A,S$ and $C,B,K$ are collinear. Let tangents to $\omega_1$ and $\omega_2$ at $A$ and $B$ meet at $Q$. $\angle QAC = \angle ABC = \angle ASK \implies C,A,S$ are collinear. $\angle QBC = \angle BAC = \angle BKS\implies C,B,K$ are collinear. So now we have $S$ and $K$ are two of those 4 points and same way we can prove $T$ and $M$ are other points. we're Done.

Thought inversions first, but unforunately don't know how inversions work. Let the two given circles be $S_1,S_2,$ with the new circle $C$ touching $S_1,S_2$ at $A,B$ respectively. Let $C$ be closer than $D$ to $Y.$ Let $CA$ intersect $S_1$ at $E,$ $CB$ intersect $S_2$ at $F,$ $DA$ intersect $S_1$ at $G$ and $DB$ intersect $S_2$ at $H.$ We claim that $EF,GH$ are always the common tangents.Note that since $C,D$ are on the radical axis of $S_1,S_2,$ $ABFE$ and $ABHG$ are cyclic quadrilaterals. Thus $\angle GEA=\angle DCA$ (famous lemma, to prove use homothety about $A$) and so $\angle DCA=\angle DBA=\angle HGA,$ so $HG$ tangent to $S_1$ Similarly, $GH$ tangent to $S_2.$ On the other hand, $\angle BHF=\angle CDB=\angle CAB=\angle CFB$ so $EF$ tangent to $S_2,$ and similarly to $S_1,$ as desired.

Let $\measuredangle$ denote a directed angle. Let $\omega_1, \omega_2$ be the original two circles and let $\omega$ be a circle that is tangent to them at $A, B$ respectively. We claim that the four distinct points $P_1, P_2, P_3, P_4$ such that $P_1P_2$ and $P_3P_4$ are each tangent to both $\omega_1$ and $\omega_2$ satisfy the problem statement. Let $C$ be an intersection of $XY$ and $\omega$. Let $A'$ be the second intersection of $AC$ with $\omega_1$ and let $B'$ be the second intersection of $BC$ with $\omega_2$. (Note that $CA$ cannot be tangent to $\omega_1$ at $A$, since then it would have to tangent to $\omega$ as well.) We claim $A'B'$ is tangent to $\omega_1$ and $\omega_2$ at $A'$ and $B'$. Because $C$ is on the radical axis of $\omega_1$ and $\omega_2$, $CA\cdot CA'=CB\cdot CB'$. Therefore, $AA'BB'$ is cyclic. So, we have $\measuredangle B'A'A=\measuredangle B'BA=\measuredangle CBA$. Similarly, $\measuredangle A'B'B=\measuredangle CAB$. Let $O$ be the radical center of the three circles. Note $OA$ is tangent to $\omega_1, \omega$ and $OB$ is tangent to $\omega_2, \omega$. So, we have $\measuredangle CAO=\measuredangle CBA=\measuredangle B'A'A$. Let $P$ be a point on $\omega_1$ not equal to $A$ or $A'$. Since $OA$ is tangent to $\omega_1$, we have $\measuredangle A'PA=\measuredangle A'AO=\measuredangle CAO=\measuredangle B'A'A$. Since $\measuredangle A'PA=\measuredangle B'A'A$, we know $A'B'$ is tangent to $\omega_1$ at $A'$. By the same logic $A'B'$ is tangent to $\omega_2$ at $B'$. Thus $A', B'\in\{P_1, P_2, P_3, P_4\}$, which completes the proof since $CA$ and $CB$ pass through $A'$ and $B'$, respectively.

Does this work? Not sure if I fakesolved. Consider the diagram as shown below. I claim that the four points $E$, $F$, $G$, and $H$ which we require are the four points which are the intersections of the two external tangents of the circles, which we call $\omega_1$ and $\omega_2$. Also, define $M$ to be the intersection of $XY$ and $GH$. Obviously $A$, $B$, $G$, and $H$ are concyclic which means that $$\measuredangle DCA=\measuredangle DBA=\measuredangle HGD$$which implies that $A$, $C$, $M$, and $G$ are concyclic. By Reim's Theorem, $CD\parallel EG$, which implies that $$\measuredangle HGA=\measuredangle DCA=\measuredangle GEA$$so that $GH$ is tangent to $\omega_1$. The final result immediately follows. $\blacksquare$

this is extremely similar to 2021 aime i 13 and actually trivializes it (angle between external tangent and line connecting centers is 30 degrees) very nice

Let $\odot (AXY)$ meets rays $AC,AD$ again at $P,Q$ and $\odot (BXY)$ meets $BC,BD$ again at $R,S.$ Since $C$ lies on radical axis of given circles, points $A,B,P,R$ are concyclic, so by Reim's theorem $\overline{PR}$ is parallel to tangent to $\odot (ABC)$ at $C.$ Thus by homothety $PR$ is tangent to $\odot (AXY),\odot (BXY),$ which yields that points $P,R$ (and similarly $Q,S$) are fixed.

I claim these points are the intersections of the common tangents of the two circles with the circles. Let $K,L$ be the intersection of one of the common tangents of circles with the circles, then we prove $AK$, $BL$, $XY$, concur on the circle $(ABCD)$. First by Monge, $AB$ meets the common tangents of the circles at $Z$, so let $ZAB$ meet $(BXY)$ at $A'$, then we know that $ZK \cdot ZL = \frac{ZK}{ZL} ZL^2 = \frac{ZK}{ZL} ZB \cdot ZA' = \frac{ZA}{ZA'} ZB \cdot ZA' = ZA \cdot ZB$, so $(KLAB)$ is cyclic. By RadAx, $KA, BL, XY$ concur at $O$. For this angle chase, we assume $K,L$ are on the same side of $O_1O_2$ as $AB$, although an identical argument works in the other case. We now let $\angle AO_1O_2 = x, \angle BO_2O_1 = y, \angle KO_1O_2 = k$, then we have $\angle AO_3B = 180 - x - y$. Now $\angle LKO = 90 - \angle OKO_1 = 90 - \angle OKA = 90 - 90 + \frac 12\angle KO_1A = \frac 12 (k - x)$. Likewise, $\angle KLO = \frac 12 (180 -k - y)$. Now $180 - \angle KLO - \angle OKL = 180 - \angle OAB - \angle OBA = \angle AOB = 90 + \frac 12 (x + y)$. Now $\frac 12 \angle AO_3B = 90 - \frac 12 (x + y) $, so for some random point $R$ on arc $AB$ we have that $ARBO$ is cyclic, so $O$ lies on the third circle, and since it also lies on $XY$ it must be one of $C$ or $D$. Repeating this argument for the other tangent forces a different concurrency point, so we are done.