6. The ellipse $C$ is symmetric to the ellipse $\frac{(x-3)^{2}}{9}+\frac{(y-2)^{2}}{4}=1$ with respect to the line $x+y=0$. The equation of the ellipse $C$ is
A. $\frac{(x+2)^{2}}{4}+\frac{(y+3)^{2}}{9}=1$
B. $\frac{(x-2)^{2}}{9}+\frac{(y-3)^{2}}{4}=1$
C. $\frac{(x+2)^{2}}{9}+\frac{(y+3)^{2}}{4}=1$
D. $\frac{(x-2)^{2}}{4}+\frac{(y-3)^{2}}{9}=1$
6. A Hint: Let $(x, y)$ be any point on the ellipse $C$, then the coordinates of the point symmetric to $(x, y)$ with respect to $x+y=0$ are $(-y, -x)$. Therefore, we have $\frac{(-y-3)^{2}}{9}+\frac{(-x-2)^{2}}{4}=1$, hence the answer is A.