4. In $\triangle A B C$, it is known that $A B=A C=8 \sqrt{3}$, and the altitude $A D$ $=8$. Then the radius of the circumcircle of $\triangle A B C$ is $(\quad$.
(A) 8
(B) 9
(C) 10
(D) 12
4.D.

As shown in Figure 4, extend $AD$ to intersect the circumcircle at point $E$, then $AE$ is the diameter. Connecting $BE$, we know that $\triangle ABE$ is a right triangle, thus we have
$$
AB^{2}=AD \cdot AE .
$$

Therefore, the radius is 12.