One, (20 points) Let $a, b$ be real numbers, and
$$
4 a^{2}+16 b^{2}-16 a b+13 a-26 b+3=0 \text {. }
$$

Find the range of $a^{2}+b^{2}$.
i. Completing the square, factoring, and letting $t=a-2 b$, we get
$$
(4 t+1)(t+3)=0 \text {. }
$$

Solving for $t$ yields $t=-3$ or $t=-\frac{1}{4}$.
When $a-2 b=-3$, we have
$$
a^{2}+b^{2}=(2 b-3)^{2}+b^{2}=5 b^{2}-12 b+9,
$$

the minimum value of which is $\frac{9}{5}$;
When $a-2 b=-\frac{1}{4}$, we have
$$
a^{2}+b^{2}=\left(2 b-\frac{1}{4}\right)^{2}+b^{2}=5 b^{2}-b+\frac{1}{16} \text {, }
$$

the minimum value of which is $\frac{1}{80}$.
In summary, the range of $a^{2}+b^{2}$ is $\left[\frac{1}{80},+\infty\right)$.