6. Point $P$ in the Cartesian coordinate system has coordinates $(a, b)$, where $a$ and $b$ satisfy
$$
\left\{\begin{array}{l}
(a+b)^{2 a}(b+a)^{2 b}=(a+b)^{14}, \\
(a-b)^{2 a+7}(a-b)^{6-2 b}=(a-b)^{11} .
\end{array}\right.
$$

Then the number of possible positions of point $P$ in the Cartesian coordinate system is ( ).
(A) $1 \sim 4$
(B) 5 or 6
(C) 7 or 8
(D) $9 \sim 12$
6.C.

From equation (1), we get
$$
(a+b)^{2 a+2 b}=(a+b)^{14} \text {. }
$$

Thus, $2 a+2 b=14$ or $a+b=0, \pm 1$.
When $a+b=0$, at least one of $a$ and $b$ is 0 or a negative number, making $(a+b)^{2 a}$ or $(b+a)^{2 b}$ undefined.
When $a+b= \pm 1$, $2 a+2 b$ is always even, so equation (1) always holds.
Therefore, $a+b=7$ or $\pm 1$.
From equation (2), we get $(a-b)^{2 a-2 b+13}=(a-b)^{11}$.
Thus, $2 a-2 b+13=11$ or $a-b=0, \pm 1$.
So, $a-b=0, \pm 1$.
Solving equations (3) and (4) together, we get
$$
\begin{array}{c}
(a, b)=(3.5,3.5),(4,3),(3,4),(0.5,0.5), \\
(1,0),(0,1),(-0.5,-0.5),(0,-1),(-1,0) .
\end{array}
$$

When $(a, b)=(3.5,3.5)$, $(a-b)^{6-2 b}$ is undefined, so there are 8 possible positions for point $P$.