2. As shown in Figure 1, quadrilateral $ABCD$ is an inscribed quadrilateral of $\odot O$, and $AC \perp BD$ at point $E$. If $AB=8, CD=6$, then the radius of $\odot O$ is ().
(A) 6
(B) $2 \sqrt{7}$
(C) 5
(D) $\sqrt{7}$
2. C.

As shown in Figure 4, draw the diameter $B F$ of $\odot O$ through point $B$, and connect $A F, D F$.

Since $B F$ is the diameter of $\odot O$, we know
$$
\begin{array}{l}
B D \perp D F . \\
\text { Since } A C \perp B D, \text { therefore, } \\
A C \parallel F D \\
\Rightarrow A F=C D=6 .
\end{array}
$$

Then $B F=\sqrt{A B^{2}+A F^{2}}=10$.
Thus $O B=5$.