5. As shown in Figure 3, in $\triangle A B C$, $\angle C=90^{\circ}, A C=\sqrt{11}$, $B C=5$, a circle is drawn with $C$ as the center and $B C$ as the radius, intersecting the extension of $B A$ at point $D$. Then the length of $A D$ is ( ).
(A) $\frac{3}{7}$
(B) $\frac{5}{7}$
(C) $\frac{7}{3}$
(D) $\frac{5}{3}$
5.C.

As shown in Figure 8, extend $AC$ to intersect the circle at $E$ and $F$, then
$$
\begin{array}{l}
A F=5+\sqrt{11}, \\
A E=5-\sqrt{11} .
\end{array}
$$

It is easy to calculate that $A B=6$. By the intersecting chords theorem $A D \cdot A B=A E \cdot A F$
we get
$$
A D=\frac{A E \cdot A F}{A B}=\frac{(5-\sqrt{11})(5+\sqrt{11})}{6}=\frac{7}{3} .
$$