Example 12. Real numbers $x, y$ satisfy $x^{2}+y^{2}-4 x$ $-2 y+5=0$. Then the value of $\frac{\sqrt{x}+y}{\sqrt{3 y-2 \sqrt{x}}}$ is ( ).
(A) 1
(B) $\frac{3}{2}+\sqrt{2}$
(C) $3+2 \sqrt{2}$
(D) $3-2 \sqrt{2}$
Solving, the given conditions can be rewritten as
$$
x^{2}-4 x+4+y^{2}-2 y+1=0 \text {. }
$$

which is $(x-2)^{2}+(y-1)^{2}=0$.
Thus, $x-2=0, y-1=0$,
which means $x=2 ; y=1$.
Therefore, the original expression $=\frac{\sqrt{2}+1}{\sqrt{3}-2 \sqrt{2}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$
$$
=3+2 \sqrt{2} \text {. }
$$

Hence, the correct choice is (C).