6. Set $R$ represents the set of points $(x, y)$ in the plane, where $x^{2}+6 x+1+y^{2}+6 y+1 \leqslant 0$, and $x^{2}+6 x+1-\left(y^{2}+\right.$ $6 y+1) \leqslant 0$. Then, the area of the plane represented by set $R$ is closest to ( ).
(A) 25
(B) 24
(C) 23
(D) 22
6.A.

From the given, we have $(x+3)^{2}+(y+3)^{2} \leqslant 16$,
$$
(x-y)(x+y+6) \leqslant 0 \text {. }
$$

The first inequality represents the interior and boundary of a circle with center $(-3,-3)$ and radius 4; the second inequality can be written as $x-y \geqslant 0$ and $x+y+6 \leqslant 0$, or $x-y \leqslant 0$ and $x+y+6 \geqslant 0$.
The area enclosed by the above figures is $8 \pi \approx 25.13$. Therefore, the area of the plane represented by the set $R$ is closest to 25.