4. Given $O$ as the origin, $\overrightarrow{O M}=(-1,1), \overrightarrow{N M}=(-5,-5)$, the set $A=\{\overrightarrow{O R}|| R N \mid=2\}$, $\overrightarrow{O P}, \overrightarrow{O Q} \in A, \overrightarrow{M P}=\lambda \overrightarrow{M Q}(\lambda \in \mathbf{R}, \lambda \neq 0)$, then $\overrightarrow{M P} \cdot \overrightarrow{M Q}=$
4. 46 .

According to the problem, points $M, P, Q$ are collinear and form a secant of the circle $(x-4)^{2}+(y-6)^{2}=4$. As shown in Figure 2, take the midpoint $A$ of $P Q$, we can get
$$
\begin{aligned}
\overrightarrow{M P} \cdot \overrightarrow{M Q} & =(\overrightarrow{M A}-\overrightarrow{P A}) \cdot(\overrightarrow{M A}+\overrightarrow{A Q}) \\
& =|M A|^{2}-|A P|^{2}=|M N|^{2}-|N A|^{2}-|A P|^{2} \\
& =|M N|^{2}-|N P|^{2}=46 .
\end{aligned}
$$