1. Given that $x=-2$ is a root of the equation $x^{2}+a x+2 b=0$. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
From the condition, we have
$$
4-2 a+2 b=0 \Rightarrow b=a-2 \text {. }
$$

Then $a^{2}+b^{2}=a^{2}+(a-2)^{2}$
$$
=2 a^{2}-4 a+4=2(a-1)^{2}+2 \geqslant 2 \text {. }
$$

When $a=1, b=-1$, the equality holds.
Therefore, the minimum value of $a^{2}+b^{2}$ is 2.