9. Given a sphere with radius 4, there are two points $A, B$ on the sphere, $A B=4 \sqrt{2}$, and the center of the sphere is $O$. If a moving point $C$ on the sphere satisfies that the dihedral angle $C-A B-O$ is $60^{\circ}$, then the radius of the circumscribed sphere of the tetrahedron $O A B C$ is $\qquad$.
9. $\frac{4 \sqrt{6}}{3}$.

It is known that the trajectory of $C$ is a major arc, and let the center of this major arc be $D$. Connect $O D$, as shown in Figure 4.
Let the midpoint of $A B$ be $E$, and connect $O E$ and $D E$. Then $\angle D E O=60^{\circ}$.
Extend $O D$, and draw a perpendicular from point $E$ to plane $O A B$, intersecting $O D$ at point $K$.
Then $K$ is the center of the circumscribed sphere of the tetrahedron $O A B C$. Thus, $O E=\frac{1}{2} A B=2 \sqrt{2}, \angle D O E=30^{\circ}$. Therefore, the required radius is $K O=\frac{4 \sqrt{2}}{\sqrt{3}}=\frac{4 \sqrt{6}}{3}$.