9. (14 points) A soldier needs to check for landmines in a circular area with a radius of $2 a \mathrm{~m}$. The effective range of the detection instrument he uses is $a \mathrm{~m}$. Find the minimum distance the soldier needs to walk starting from a point $A$ on the boundary of the circle, to complete the inspection and return to the starting point.

---

The translation maintains the original format and line breaks as requested.
$=9 \cdot\left(\frac{4 \pi}{3}+2 \sqrt{3}\right) a$.
First, find the shortest path $\Gamma$ for a soldier to start from $A$, inspect all points on the boundary of the circle $D(O, 2a)$, and return to $A$.
Below, we use proof by contradiction to show:
(1) Any line segment connecting two points on $\Gamma$ lies within the region enclosed by $\Gamma$ (including the boundary), i.e., $\Gamma$ is convex;
(2) $\Gamma$ does not intersect the interior of the circle $D(O, a)$.
(1) Otherwise, suppose $B, C \in \Gamma$, and the line segment $BC$ lies outside the region enclosed by $\Gamma$ (as shown in Figure 2).

Replace the curve between $B$ and $C$ on $\Gamma$ with the line segment $BC$ to obtain a new closed curve $\Gamma'$. Then the curve $\Gamma$ lies within the region enclosed by $\Gamma'$ (including the boundary).
For any point $E$ on the boundary of the circle $D(O, 2a)$, let the soldier inspect $E$ at point $E'$ on $\Gamma$, then $|EE'| \leq a$.
Take the intersection of the line segment $EE'$ with $\Gamma'$ as $E''$, then
$|EE''| \leq |EE'| \leq a$.
Thus, the soldier can also inspect all points on the boundary of the circle $D(O, 2a)$ by following $\Gamma'$. However, the length of $\Gamma'$ is less than the length of $\Gamma$, which is a contradiction.
(2) Otherwise, suppose $B, C \in \Gamma \cap D(O, a)$, and the curve $\overparen{BC}$ between $B$ and $C$ on $\Gamma$ lies inside the circle $D(O, a)$ (as shown in Figure 3).
Draw a perpendicular from the center $O$ to $BC$ intersecting the circle $D(O, 2a)$ at point $E$, where $E$ is on the same side of the line $BC$ as the curve $\overparen{BC}$. Let the intersection of the line segment $OE$ with the circle $D(O, a)$ be $E'$.

By the convexity of $\Gamma$, the curve $\overparen{BC}$ and the rest of $\Gamma$ lie on opposite sides of the line $BC$. Then
$$
\min_{E'' \in \Gamma} |E''E| = \min_{E'' \in \overparen{BC}} |E''E| > |E'E| = a,
$$

which means the soldier cannot inspect point $E$ by following $\Gamma$, which is a contradiction.
From (1) and (2), we know that $\Gamma$ is a closed curve containing point $A$ and enclosing the circle $D(O, a)$. The minimum length of $\Gamma$ is $\left(\frac{4 \pi}{3} + 2 \sqrt{3}\right) a$ (imagine $\Gamma$ as a rope wrapped around the circle $D(O, a)$; when the rope is pulled tight from point $A$, the shortest length of the rope is $\left(\frac{4 \pi}{3} + 2 \sqrt{3}\right) a$).

It is easy to prove that when $\Gamma_{\min} = \left(\frac{4 \pi}{3} + 2 \sqrt{3}\right) a$, the soldier can inspect all points inside the circle $D(O, 2a)$ by following $\Gamma$.