Notice that $AX$ is the perpendicular bisector of $BH$, then $\angle HAM=\frac{\angle HAB}{2}=\angle HGM$ and $AGHM$ cyclic. With this we have $XC\cdot XC'=XM\cdot XA$ where $C'=XA\cap k$, then $(XC)(XC+2AC)=(XC+CM)(XC+AC) \implies XC^2+2(AC)(XC)=XC^2+XC(AC+CM)+(AC)(CM) \implies 2(AC)(XC)=(XC)(\frac{3}{2}\cdot AC)+(AC)(CM)\implies \frac{(AC)(XC)}{2}=(AC)(CM)\implies XC=2CM=CA$

See my solution on my Youtube channel here: https://www.youtube.com/watch?v=cZT2wG50w-8

Since $AC$ is the perpendicular bisector of $BH$, then $BX=HX$ and $MX$ is the angle bisector of $\angle BXG$ in $\triangle BXG$. Also, since $\overset{\huge\frown}{BC}= \overset{\huge\frown}{CH}$, $CG$ is also an angle bisector of $\triangle GMX$, we then have (denote $D=XA\cap k, D\ne C$) \begin{align*} XC(XC+2AC)&= XC\cdot SD \\ &= XH\cdot XG \\ &= XB\cdot XG \\ &= \frac{MB\cdot XG}{MG} \cdot \frac{XC\cdot MG}{MC} \\ &= \frac{MB\cdot XC\cdot MG}{MG\cdot MC} \cdot \frac{XC\cdot MG}{MC} \\ &= \frac{MB\cdot MG\cdot XC^2}{MC^2} \\ &= \frac{MD\cdot MC \cdot XC^2}{MC^2} \\ &= \frac{\frac{3}{2}AC\cdot XC^2}{\frac{1}{2}AC} \\ &= 3XC^2. \end{align*}Hence, $AD=XC. \blacksquare$

Let $D$ be the antipode of $C$ wrt the circle $k$ and $M$ be the midpoint of $AC$. Then $DBCH$ is a kite so $(D,C;B,H)=-1$. Consider projection centered at $G$ onto the line $AC$ we get $(D,C;M,X)=-1$. From here we obtain that $XC=R$ where $R$ is the radius of the circle $k$. On other hand $AC=R$ so we are done.