smc

Construct the circle that passes through $A$, $O$, and $C$, centered at $X$. Also notice that $\overline{OA}$ and $\overline{OC}$ are the angle bisectors of angle $\angle BAC$ and $\angle BCA$ respectively. We then deduce $\angle AOC=120^{\circ }$. Consider another point $M$ on Circle $X$ opposite to point $O$. As $AOCM$ is an inscribed quadrilateral of Circle $X$, $\angle AMC=180^{\circ }-120^{\circ }=60^{\circ }$. Afterward, deduce that $\angle AXC=2\cdot \angle AMC=120^{\circ }$. By the Cosine Rule, we have the equation: (where $r$ is the radius of circle $X$) $2r^2(1-\cos (120^{\circ }))=6^2$ $r^2=12$ The area is therefore $\pi r^2=\boxed{12\pi }$.