3. Given two equations about $x$: $x^{2}+2 b x+a$ $=0$ and $x^{2}+a x+2 b=0$ have exactly one common root. Then the minimum value of $a^{2}+b^{2}$ is $(\quad)$.
(A) $\frac{1}{5}$
(B) $\frac{2}{5}$
(C) $\frac{3}{5}$
(D) $\frac{4}{5}$
3. A.

Let the common root of the two equations be $x=t$, then we have $t^{2}+2 b t+a=0, t^{2}+a t+2 b=0$.
Subtracting the two equations yields
$$
(2 b-a) t+a-2 b=0 .
$$

Given that $a-2 b \neq 0$, we have $t=1$.
Substituting $t=1$ into the equation gives $1+2 b+a=0$.
Thus, $a^{2}+b^{2}=(-1-2 b)^{2}+b^{2}$
$$
=5 b^{2}+4 b+1=5\left(b+\frac{2}{5}\right)^{2}+\frac{1}{5} \text {. }
$$

Therefore, when $b=-\frac{2}{5}$, the minimum value is $\frac{1}{5}$.