4. As shown in Figure 2, let $P$ and $Q$ be moving points on two concentric circles (with radii 6 and 4, respectively). When $P$ and $Q$ move along the circles, the area of the region formed by the midpoint $M$ of line segment $PQ$ is $\qquad$
4. $24 \pi$.

Draw $O H \perp P Q$ through the center $O$, with the foot of the perpendicular at $H$. Let $O P=R$, $O Q=r, O M=m, O H=h$. Then
$$
\begin{array}{l}
P M=\sqrt{R^{2}-h^{2}}-\sqrt{m^{2}-h^{2}}, \\
Q M=\sqrt{r^{2}-h^{2}}+\sqrt{m^{2}-h^{2}} .
\end{array}
$$

Since $M$ is the midpoint of $P Q$, we have
$$
\begin{array}{l}
P M=Q M \\
\Rightarrow \sqrt{R^{2}-h^{2}}-\sqrt{r^{2}-h^{2}}=2 \sqrt{m^{2}-h^{2}} \\
\Rightarrow \frac{R^{2}-r^{2}}{\sqrt{R^{2}-h^{2}}+\sqrt{r^{2}-h^{2}}}=2 \sqrt{m^{2}-h^{2}} \\
\Rightarrow \sqrt{R^{2}-h^{2}}+\sqrt{r^{2}-h^{2}}=\frac{R^{2}-r^{2}}{2 \sqrt{m^{2}-h^{2}}} .
\end{array}
$$
Adding (1) and (2) gives
$$
\begin{array}{l}
2 \sqrt{R^{2}-h^{2}}=2 \sqrt{m^{2}-h^{2}}+\frac{R^{2}-r^{2}}{2 \sqrt{m^{2}-h^{2}}} \\
\Rightarrow 4\left(R^{2}-h^{2}\right) \\
\quad=4\left(m^{2}-h^{2}\right)+2\left(R^{2}-r^{2}\right)+\frac{\left(R^{2}-r^{2}\right)^{2}}{4\left(m^{2}-h^{2}\right)} \\
\Rightarrow 2\left(R^{2}+r^{2}-2 m^{2}\right)=\frac{\left(R^{2}-r^{2}\right)^{2}}{4\left(m^{2}-h^{2}\right)} \\
\Rightarrow h^{2}=m^{2}-\frac{\left(R^{2}-r^{2}\right)^{2}}{8\left(R^{2}+r^{2}-2 m^{2}\right)} .
\end{array}
$$

Substituting $R=6, r=4$ into equation (3) gives
$$
h^{2}=m^{2}-\frac{25}{26-m^{2}} \text {. }
$$

From $m^{2}-\frac{25}{26-m^{2}} \geqslant 0$, we solve to get
$1 \leqslant m^{2} \leqslant 25$ or $m^{2}>26$.
If $m^{2}>26$, then $h^{2}>m^{2}>26$, which contradicts $h \leqslant r=4$.
Therefore, $1 \leqslant m \leqslant 5$.
From the previous derivation, we know that for any point $M$ within the annular region with center $O$ and radius $1 \leqslant m \leqslant 5$, there exists a line segment $P Q$ with a distance of $h(m)$ from the center, such that points $P$ and $Q$ lie on the two circles, and the midpoint of $P Q$ is $M$.
Thus, the area of the region formed by point $M$ is
$$
\pi\left(5^{2}-1^{2}\right)=24 \pi \text {. }
$$