jmc

We can begin by using the circumference to solve for the radius of the circle. If the circumference is $12\pi$, then $2\pi r=12\pi$ which implies $r=6$. Now, we can draw in the radius $TZ$ as shown:

We know that $TX=TZ$, since both are radii of length 6. We are given $\angle TXZ=60^{\circ }$, so $\angle TZX=60^{\circ }$, and triangle $TXZ$ is equilateral. Thus, $TX=TZ=XZ=\boxed{6}$.