Take a cross-section of a plane through the centers of $\Omega$ and $\Gamma$, call them $O_{1}$ and $O_{2}$, respectively. The resulting figure is two circles, one of radius 4 and center $O_{1}$, and the other with radius 2 and center $O_{2}$ on the circle of radius 4 . Let these two circles intersect at $A$ and $B$. Note that $\overline{A B}$ is a diameter of the desired circle, so we will find $A B$. Focus on triangle $O_{1} O_{2} A$. The sides of this triangle are $O_{1} O_{2}=O_{1} A=4$ and $O_{2} A=2$. The height from $O_{1}$ to $A O_{2}$ is $\sqrt{4^{2}-1^{2}}=\sqrt{15}$, and because $O_{1} O_{2}=2 \cdot A O_{2}$, the height from $A$ to $O_{1} O_{2}$ is $\frac{\sqrt{15}}{2}$. Then the distance $A B$ is two times this, or $\sqrt{15}$. Thus, the circumference of the desired circle is $\pi \sqrt{15}$.