6. In the Cartesian coordinate system $x O y$, the circle $C_{1}: x^{2}+y^{2}-a=0$ is symmetric to the circle $C_{2}: x^{2}+y^{2}+2 x-2 a y+3=0$ with respect to the line $l$. Then the equation of the line $l$ is $\qquad$.
Answer: $2 x-4 y+5=0$.
Solution: The standard equations of $C_{1}, C_{2}$ are respectively
$$
C_{1}: x^{2}+y^{2}=a, \quad C_{2}:(x+1)^{2}+(y-a)^{2}=a^{2}-2 .
$$

Since the two circles are symmetric with respect to the line $l$, their radii are equal. Therefore, $a=a^{2}-2>0$, solving this gives $a=2$. Hence, the centers of $C_{1}, C_{2}$ are $O_{1}(0,0), O_{2}(-1,2)$ respectively. The line $l$ is the perpendicular bisector of the segment $O_{1} O_{2}$, passing through the midpoint $M\left(-\frac{1}{2}, 1\right)$ of $O_{1} O_{2}$. From this, we can derive the equation of the line $l$ as $2 x-4 y+5=0$.