smc

First, note the triangle $ABC$ is equilateral. Next, notice that since the arc $BC$ has length 12, it follows that we can find the radius of the sector centered at $A$. $\frac{1}{6}(2\pi )AB=12⟹AB=36/\pi$. Next, connect the center of the circle to side $AB$, and call this length $r$, and call the foot $M$. Since $ABC$ is equilateral, it follows that $MB=18/\pi$, and $OA$ (where O is the center of the circle) is $36/\pi -r$. By the Pythagorean Theorem, you get $r^2+(18/\pi {)}^2=(36/\pi -r{)}^2⟹r=27/2\pi$. Finally, we see that the circumference is $2\pi \cdot 27/2\pi =\boxed{27}$.