9. In $\triangle A B C$, $a=BC, b=CA, c=AB$. If $a^{2}+b^{2}=6, \cos (A-B) \cos C=\frac{2}{3}$, find the circumradius of $\triangle A B C$.
$$
\begin{array}{l}
\cos (A-B) \cos (A+B)=\frac{1}{2}(\cos 2 A+\cos 2 B)=\left(\cos ^{2} A+\cos ^{2} B\right)-1 \\
=-\cos (A-B) \cos C=-\frac{2}{3} \Rightarrow \cos ^{2} A+\cos ^{2} B=\frac{1}{3} \Rightarrow \sin ^{2} A+\sin ^{2} B=\frac{5}{3} .
\end{array}
$$

Let the circumradius of $\triangle A B C$ be $R$, then $a^{2}+b^{2}=4 R^{2}\left(\sin ^{2} A+\sin ^{2} B\right)$
$$
\Rightarrow \frac{20}{3} R^{2}=6 \Rightarrow R^{2}=\frac{9}{10} \Rightarrow R=\frac{3 \sqrt{10}}{10} \text {. }
$$