5. As shown in Figure 1, in the Cartesian coordinate system, there is a rectangle $A O B C$. The inverse proportion function $y=\frac{k}{x}$ passes through the intersection point $G$ of the diagonals of rectangle $A O B C$. A circle with radius $4-2 \sqrt{2}$ is inscribed in $\triangle A B C$. Then the minimum value of $k$ is ().
(A) 4
(B) $4 \sqrt{2}$
(C) 2
(D) $2 \sqrt{2}$
5. A.

Let $G(a, b)$, and denote the radius of the circle as $r$. Then
$$
\begin{aligned}
& A C=2 a, B C=2 b, \\
& r=\frac{1}{2}(A C+B C-A B) \\
= & a+b-\sqrt{a^{2}+b^{2}} \\
= & \frac{2 a b}{a+b+\sqrt{a^{2}+b^{2}}} \\
\leqslant & \frac{2 a b}{2 \sqrt{a b}+\sqrt{2 a b}} \\
= & (2-\sqrt{2}) \sqrt{a b} .
\end{aligned}
$$

Thus, $k=a b \geqslant\left(\frac{r}{2-\sqrt{2}}\right)^{2}=4$.
The equality holds if and only if $a=b=2$.