5. Given parallel lines $x+y=2 a-1$ and the family of concentric circles $x^{2}+y^{2}=a^{2}+2 a-3$ intersect at $\left(x_{0}, y_{0}\right)$. When $x_{0} y_{0}$ takes the minimum value, the value of $a$ is ( ).
(A) 1
(B) -3
(C) $2-\frac{\sqrt{2}}{2}$
(D) $2+\frac{\sqrt{2}}{2}$
5. (C).

From the inequality $(x+y)^{2} \leqslant 2\left(x^{2}+y^{2}\right)$, we get
$$
(2 a-1)^{2} \leqslant 2\left(a^{2}+2 a-3\right),
$$

which simplifies to $2 a^{2}-8 a+7 \leqslant 0$.
Solving this, we get $2-\frac{\sqrt{2}}{2} \leqslant a \leqslant 2+\frac{\sqrt{2}}{2}$.
When $x=y$, $a$ can take the endpoints.
$$
\text { Also, } \begin{aligned}
x_{0} y_{0} & =\frac{1}{2}\left[\left(x_{0}+y_{0}\right)^{2}-\left(x_{0}^{2}+y_{0}^{2}\right)\right] \\
& =\frac{1}{2}\left(3 a^{2}-6 a+4\right) \\
& =\frac{3}{2}(a-1)^{2}+\frac{1}{2} .
\end{aligned}
$$

When $a=2-\frac{\sqrt{2}}{2}$, $x_{0} y_{0}$ takes the minimum value $\left(\frac{3-\sqrt{2}}{2}\right)^{2}$. Therefore, the answer is
(C).