Call the orthocenter of $AXY$ $H'$. Like the previous solution, it can be shown that $\frac{MH}{HY}$ is a constant number. Therefore there exists a constant spiral similarity with the center $M$ that maps $H$ to $Y$. So since the $H$ moves on a circle, $Y$ also moves on a circle like $\omega'$. If we name the antipode of $C$ with reference to $\omega$ $C'$, simply it's possible to observe that $\omega'$ also passes through points $C$ and $C'$. Assume that $E$ is the intersection of $H'Y$ and $\omega'$. Note that $$\measuredangle CC'E=\measuredangle CYE=\measuredangle AYH'= 90^\circ-\measuredangle XAY=90^\circ-\measuredangle BAC$$So $E$ is a constant point. Similarly, if we define $F$ as the intersection place of $H'X$ and the circumscribed circle of the triangle $BXB'$ ($B'$ antipode of $B$ wrt $\omega$), $F$ is a constant point. Also, it's obvious that $\measuredangle EH'F=\measuredangle BAC$ so $H'$ moves on a circle that also passes through $E$ and $F$. [asy][asy] size(350); pen p=linewidth(1.3); pen qqwuqq = rgb(0,0.39215686274509803,0); pen t=RGB(229,239,229); pair A, B, C, M, H, B2, C2, X, Y, H2, E, F, Ob, Oc, rx, ry, d1, d2; real r=4/Tan(68.95)**2; real perp=0.1; A=dir(107.8); B=dir(199.1); C=dir(-23); d1=dir(55); d2=dir(110); B2=-B; C2=-C; M=(B+C)/2; H=2M+A; rx=(B2-M)/(C-M); ry=(C2-M)/(B-M); X=M+rx*(H-M); Y=M+ry*(H-M); E=C+rx*(B2-C); F=B+ry*(C2-B); Ob=M*(1+rx); Oc=M*(1+ry); H2=A+r*(-M-A); pair BC=dir(C-B); pair i=dir(90); filldraw(B--(B+perp*BC)--(B+perp*BC*(1+i))--(B+perp*BC*i)--cycle,p+t); draw(B--(B+perp*BC)--(B+perp*BC*(1+i))--(B+perp*BC*i)--cycle,p+qqwuqq); filldraw(C--(C-perp*BC)--(C+perp*BC*(-1+i))--(C+perp*BC*i)--cycle,p+t); draw(C--(C-perp*BC)--(C+perp*BC*(-1+i))--(C+perp*BC*i)--cycle,p+qqwuqq); pair XY=dir(X-Y); filldraw(H--(H+perp*XY)--(H+perp*XY*(1+i))--(H+perp*XY*i)--cycle,p+t); draw(H--(H+perp*XY)--(H+perp*XY*(1+i))--(H+perp*XY*i)--cycle,p+qqwuqq); draw(A--B--C--cycle,p); draw(B--C2,p);draw(C--B2,p); draw(M--H--A,p); draw(X--Y,p); draw(E--H2--X,p); dot("$A$",A,N); dot("$B$",B,W); dot("$M$",M,S); dot("$C$",C,E); dot("$B'$",B2,d1); dot("$C'$",C2,d2); dot("$H$",H,C); dot("$X$",X,d2); dot("$Y$",Y,d1); dot("$H'$",H2,N); dot("$E$",E,NE); dot("$F$",F,d2); draw(unitcircle,p); draw(circle(Ob,abs(rx)),p); draw(circle(Oc,abs(ry)),p); draw(circle(-M*r,1-r),p+linetype(new real[] {2,3})); clip((-2,-1.01)--(-2,1.1)--(2,1.1)--(2,-1.01)--cycle); [/asy][/asy]

Call the center of $\omega$ $O$ and call the orthocenter of $AXY$ $H'$ and call the circumcenter of $AXY$ $O'$. Note that if $B'$ is foot of altitude of $B$ with reference to $ABC$ and $C'$ is the foot of altitude of $C$ with reference to $ABC$, then $BCB'C'$ is inscribed to a circle $\Gamma$ that its center is $M$ and $X$ is on $BB'$—a chord of $\Gamma$; and $Y$ is on $CC'$—its another chord. And because $H$ is the perpendicular foot of $M$ with reference to $XY$, due to the Butterfly theorem we have $HX=HY$. So, $O'$ lies on $HM$. Lemma: In a non-isosceles triangle $ABC$ the intersection of angle bisector (internal or external) of $A$ and perpendicular bisector of $BC$ lies on the circumcirle of $ABC$. Proof: Consider The intersection of the angle bisector and the circle $X$ and $\measuredangle BAX=\measuredangle XAC\implies\overset{\frown}{BX}=\overset{\frown}{CX}\implies BX=CX$ so $X$ lies on the perpendicular bisector of $BC$. Due to the preceding lemma, if the internal angle bisector of $\widehat{BAC}$ intersects $OM$ and $HM$ at $T$ and $L$ respectively and the external angle bisector of $\widehat{BAC}$ intersects $OM$ and $HM$ at $T'$ and $K$ then $T,T'\in\omega$ and $AKXLY$ is concyclic. So, $O'K=O'L$. So $\overrightarrow{O'K}+\overrightarrow{O'L}=\overrightarrow{0}$ and We have: $$\overrightarrow{OH'}=\overrightarrow{OA}+\overrightarrow{AH'}=\overrightarrow{OA}+2\overrightarrow{O'H}=\overrightarrow{OA}+2\overrightarrow{OH}-2\overrightarrow{OO'}=3\overrightarrow{OA}+4\overrightarrow{OM}-\overrightarrow{OK}-\overrightarrow{OL}$$And note that $AH\parallel T'M$ and due to Intercept theorem: $$\frac{KA}{KT'}=\frac{AH}{T'M}\implies \overrightarrow{OK}=\frac{T'M}{T'M-AH}\overrightarrow{OA}-\frac{AH}{T'M-AH}\overrightarrow{OT'}=\frac{T'M}{TM}\overrightarrow{OA}-\frac{AH}{TM}\overrightarrow{OT'}$$Again noting that $AH\parallel TM$ and using Intercept theorem we have: $$\frac{LA}{LT}=\frac{AH}{TM}\implies \overrightarrow{OL}=\frac{TM}{TM+AH}\overrightarrow{OA}+\frac{AH}{TM+AH}\overrightarrow{OT}=\frac{TM}{T'M}\overrightarrow{OA}+\frac{AH}{T'M}\overrightarrow{OT}$$These three equations imply that there exist a vector $\overrightarrow v$ and a constant number $c$ such that $\overrightarrow{OH'}=\overrightarrow{v}+c\overrightarrow{OA}$ (because $T$ and $T'$ are mid-arc of $BC$ and are constant and $M$ and $O$ are constant too.) So if we define point $V$ such that $\overrightarrow{v}=\overrightarrow{OV}$ then $H'V=cR$ where $R$ is radius of $\omega$ and we are done$\blacksquare$ [asy][asy] size(350); pen p=linewidth(1.7); pen qqwuqq = rgb(0,0.39215686274509803,0); pen t=RGB(229,239,229); pair A, B, C, M, H, B2, C2, X, Y, H2, T, T2, K, L, O2, V, rx, ry, d1, d2; pen m,a,n; m=p+RGB(233,219,97); a=p+RGB(74,84,89); n=p+RGB(130,121,126); real r=4/Tan(68.95)**2; real perp=0.1; A=dir(107.8); B=dir(199.1); C=dir(-23); T=dir(-91.95);T2=-T; d1=dir(55); d2=d1**2; M=(B+C)/2; real OM=abs(M); real MT=1-OM; real MT2=1+OM; H=2M+A; rx=(-B-M)/(C-M); ry=(-C-M)/(B-M); X=M+rx*(H-M); Y=M+ry*(H-M); B2=A+M/B*(B-A); C2=A+M/C*(C-A); H2=A+r*(-M-A); K=(MT2*A-2*OM*T2)/MT; L=(MT*A+2*OM*T)/MT2; O2=(K+L)/2; V=-M*r; draw(A--B--C--cycle,p); draw(A--H,m); draw(K--M,m); draw(A--T--M,a); draw(X--Y,p); draw(M--T2--K,n); draw(A--H2,p); dot("$O$",(0,0),E); dot("$A$",A,N); dot("$B$",B,W); dot("$B'$",B2,W); dot("$C'$",C2,NW); dot("$M$",M,SE); dot("$C$",C,E); dot("$H$",H,C); dot("$X$",X,d2); dot("$Y$",Y,d1); dot("$H'$",H2,dir(65)); dot("$T$",T,S); dot("$T'$",T2,N); dot("$K$",K,N); dot("$L$",L,W); dot("$O'$",O2,SW); dot("$V$",V,E); draw(unitcircle,p); draw(circle(V,1-r),p+linetype(new real[] {2,3})); shipout(bbox(RGB(255,255,255),Fill)); [/asy][/asy]