5. If $a>1, b>1$, then the minimum value of $\frac{a^{2}+b^{2}}{a+b-2}$ is ( ).
(A) 4
(B) 6
(C) 8
(D) does not exist
5.A.
$$
\begin{array}{l}
\frac{a^{2}+b^{2}}{a+b-2} \geqslant \frac{2\left(\frac{a+b}{2}\right)^{2}}{a+b-2}=\frac{1}{2} \cdot \frac{(a+b-2+2)^{2}}{a+b-2} \\
=\frac{1}{2}\left[(a+b-2)+4+\frac{4}{a+b-2}\right] \\
\geqslant \frac{1}{2}\left[4+2 \sqrt{(a+b-2) \frac{4}{a+b-2}}\right]=4 .
\end{array}
$$

Equality holds if and only if \(a=b=2\).