$$
\begin{array}{l}
A=\left\{(x, y) \mid(x-2)^{2}+(y-1)^{2} \leqslant 1\right\}, \\
B=\{(x, y)|2| x-1|+| y-1 \mid \leqslant a\}, \\
A \subseteq B .
\end{array}
$$

Then the range of the real number $a$ is $\qquad$
$-1 .[2+\sqrt{5},+\infty)$.
From the graph, if $A \subseteq B$, then $a>0$, and the distance from the point $(2,1)$ to the line $2 x+y-3-a=0$ is not less than 1, i.e., $\frac{|2-a|}{\sqrt{5}} \geqslant 1 \Rightarrow a \geqslant 2+\sqrt{5}$.
Therefore, $a \in[2+\sqrt{5},+\infty)$.