6. A circle with radius $x$ is tangent to (at least one side) of a square with a side length of 1, and rolls inside it. After rolling around the square once, let the area of the region not covered by $\odot O$ be $S$. The value of $x$ that minimizes $S$ is $(\quad)$.
(A) $4-\pi$
(B) $\frac{1}{4}$
(C) $\frac{4}{20-\pi}$
(D) $\frac{1}{2}$
6. (C).

When $0 \leqslant x < \frac{1}{4}$, the area of the 4 right angles and the small central square region is $S$, then we have
$$
S=(20-\pi)\left(x-\frac{4}{20-\pi}\right)^{2}+\frac{4-\pi}{20-\pi} .
$$

When $\frac{1}{4} \leqslant x \leqslant \frac{1}{2}$, the area of the 4 corner regions is $S$, then we have
$$
S=(4-\pi) x^{2} \text {. }
$$

Thus, when $x=\frac{4}{20-\pi}$, $S$ reaches its minimum value, which is
$$
\frac{4-\pi}{20-\pi} .
$$