1. Given the set $A=\left\{(x, y) \mid(x-2)^{2}+(y-1)^{2} \leqslant 1\right\}, B=\{(x, y)|2| x-1|+| y-1 \mid \leqslant a\}, A \subseteq B$, then the range of the real number $a$ is $\qquad$ .
Solve:
$$
2|x-1|+|y-1|=\left\{\begin{array}{l}
2 x+y-3, x, y \geqslant 1 \\
2 x-y-1, x \geqslant 1, y<1 \\
-2 x+y+1, x<1, y \geqslant 1 \\
-2 x-y+3, x, y<1
\end{array}\right.
$$

As shown in the figure, the distance $d$ from point $Q(2,1)$ to the line $2 x+y-3=a$ is $d=\frac{|a-2|}{\sqrt{5}} \geqslant 1 \Rightarrow a \geqslant \sqrt{5}+2$, so the range of the real number $a$ is $[\sqrt{5}+2,+\infty)$.