5. The set
$$
A=\left\{\left.(x, y)\left|(x-t)^{2}+y^{2} \leqslant\left(1-\frac{t}{2}\right)^{2},\right| t \right\rvert\, \leqslant 2\right\}
$$

represents a plane region with an area of $\qquad$ .
5. $4 \sqrt{3}+\frac{8 \pi}{3}$.

When $t=2$, the set $A=\{(2,0)\}$;
When $t=-2$, the set
$$
A=\left\{(x, y) \mid(x+2)^{2}+y^{2} \leqslant 4\right\} \text {. }
$$

Thus, the required region is the closed area formed by drawing two tangents from the point $(2,0)$ to the circle $(x+2)^{2}+$ $y^{2}=4$ (as shown in Figure 2), and its area is $4 \sqrt{3}+\frac{8 \pi}{3}$.