3. Find the extremum of the function $y=\sqrt{4 a^{2}+x^{2}}+\sqrt{(x-a)^{2}+a^{2}}(a>0)$.
3. Given that the function contains two radicals, it is natural to consider constructing corresponding complex numbers and studying them through their modulus.

Let $z_{1}=x+2 a \mathrm{i}, z_{2}=a-x+a \mathrm{i}$, then
$$
y=\left|z_{1}\right|+\left|z_{2}\right| \geqslant\left|z_{1}+z_{2}\right|=|a+3 a i|=\sqrt{10} a \text {. }
$$

Equality holds if and only if the principal arguments of $z_{1}$ and $z_{2}$ are equal, i.e., $\frac{2 a}{x}=\frac{a}{a-x}$, which gives $x=\frac{2 a}{3}$.
Thus, $y=\sqrt{4 a^{2}+x^{2}}+\sqrt{(x-a)^{2}+a^{2}}$ achieves its minimum value $\sqrt{10} a$ when $x=\frac{2 a}{3}$.