4. (4th "Hope Cup" Invitational Competition Question) Real numbers $x, y$ satisfy the equation $x^{2}+y^{2}-6 x=0$, then the range of $\sqrt{2 x^{2}+y^{2}-4 x+5}$ is ( ).
A. $[\sqrt{5}, \sqrt{53}]$
B. $[5,53]$
C. $[4,+\infty)$
D. $[2, \sqrt{53}]$
4. A. Reason: From $x^{2}+y^{2}-6 x=0$, i.e., $(x-3)^{2}+y^{2}=9$, we know that $0 \leqslant x \leqslant 6$. Therefore, $u=\sqrt{2 x^{2}+y^{2}-4 x+5}=$ $\sqrt{x^{2}+2 x+5}=\sqrt{(x+1)^{2}+4}$ is an increasing function when $0 \leqslant x \leqslant 6$. When $x=0$, $u_{\text {min }}=\sqrt{5}$; when $x=6$, $u_{\text {max }}=\sqrt{53}$.