9. Given the figure
$$
\begin{array}{l}
C_{1}:(x+3)^{2}+y^{2}=4, \\
C_{2}: x^{2}+(y-5)^{2}=4,
\end{array}
$$

There are infinitely many pairs of perpendicular lines $l_{1} 、 l_{2}$ passing through a point $P$ in the plane, which intersect circles $C_{1}$ and $C_{2}$ respectively, and the lengths of the chords intercepted by circles $C_{1}$ and $C_{2}$ are equal. Then the coordinates of point $P$ are $\qquad$
9. $P(1,1)$ and $P(-4,4)$.

Let $P(a, b)$, the equations of lines $l_{1}$ and $l_{2}$ are respectively
$$
y-b=k(x-a), y-b=-\frac{1}{k}(x-a),
$$

i.e., $k x-y-k a+b=0, x+k y-a-b k=0$.
It is easy to see that the distance from the center $C_{1}(-3,0)$ to $l_{1}$ is equal to the distance from the center $C_{2}(0,5)$ to $l_{2}$. Thus,
$$
\frac{|-3 k-a k+b|}{\sqrt{k^{2}+1}}=\frac{|5 k-a-b k|}{\sqrt{k^{2}+1}},
$$

which simplifies to $|(3+a) k-b|=|(5-b) k-a|$.
This equation holds for infinitely many $k$, so
$$
\left\{\begin{array} { l } 
{ 3 + a = 5 - b , } \\
{ b = a }
\end{array} \text { or } \left\{\begin{array}{l}
3+a=b-5, \\
b=-5 .
\end{array}\right.\right.
$$
Therefore, $P(1,1)$ and $P(-4,4)$.