12. Given the sets $A=\left\{(x, y) \mid(x-2)^{2}+(y+3)^{2} \leqslant 4\right\}, B=$ $\left\{(x, y) \left\lvert\,(x-1)^{2}+(y-a)^{2} \leqslant \frac{1}{4}\right.\right\}$, and $A \cap B=B$, the range of the real number $a$ is . $\qquad$
12. $-3-\frac{\sqrt{5}}{2} \leqslant a \leqslant-3+\frac{\sqrt{5}}{2}$.

From $A \cap B=B$ we know $B \subseteq A$, then $\sqrt{(2-1)^{2}+(-3-a)^{2}} \leqslant 2-\frac{1}{2}$, which can be solved to get.