4. In $\triangle A B C$, $A B=15, A C=13$, and the altitude $A D=12$. Let the radius of the circle that can completely cover $\triangle A B C$ be $R$. Then the minimum value of $R$ is $\qquad$
4. $\frac{65}{8}$ or $\frac{15}{2}$.

As shown in figure (a), if $D$ is on $BC$, by the Pythagorean theorem, we get
$$
\begin{array}{l}
B D=\sqrt{A B^{2}-A D^{2}}=\sqrt{15^{2}-12^{2}}=9, \\
D C=\sqrt{A C^{2}-A D^{2}}=\sqrt{13^{2}-12^{2}}=5 . \\
\therefore B C=14 . \\
\because B C^{2}+A C^{2}=14^{2}+13^{2}>15^{2}=A B^{2},
\end{array}
$$
$\therefore \triangle A B C$ is an acute triangle. In this case, the smallest circle that can completely cover $\triangle A B C$ is the circumcircle of $\triangle A B C$.
Thus, $R=\frac{A B \cdot A C}{2 A D}=\frac{65}{8}$.
As shown in figure (b), if $D$ is on the extension of $BC$, then $B C=4$, and $\triangle A B C$ is an obtuse triangle. In this case, the smallest circle that can completely cover $\triangle A B C$ is the circle with the largest side $A B$ as its diameter, $R=$ $\frac{1}{2} A B=\frac{15}{2}$.