4. Given point $P(-2,5)$ is on
$\odot C: x^{2}+y^{2}-2 x-2 y-23=0$
, and line $l: 3 x+4 y+8=0$ intersects $\odot C$ at points $A$ and $B$. Then $\overrightarrow{A B} \cdot \overrightarrow{B C}=$ $\qquad$
4. -32 .

From the given, we have $\odot C:(x-1)^{2}+(y-1)^{2}=5^{2}$, the distance from the center $C$ to the line $l$ is 3.
$$
\begin{array}{l}
\text { Then }|\overrightarrow{B C}|=5,|\overrightarrow{A B}|=8, \\
\cos \angle A B C=\frac{4}{5} \text {. } \\
\text { Therefore, } \overrightarrow{A B} \cdot \overrightarrow{B C}=|\overrightarrow{A B}||\overrightarrow{B C}| \cos (\pi-\angle A B C) \\
=-32 \text {. }
\end{array}
$$