$13 \cdot 28$ If real numbers $x, y$ satisfy the condition $2 x^{2}-6 x+y^{2}=0$, then the maximum value of $x^{2}+y^{2}+2 x$ is
(A) 14.
(B) 15.
(C) 16.
(D) Cannot be determined.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1991)
［Solution］Given $y^{2}=-2 x^{2}+6 x$, thus we have $x^{2}+y^{2}+2 x=x^{2}-2 x^{2}+6 x+2 x=-(x-4)^{2}+16$. Also, since $y^{2} \geqslant 0$, we have $-2 x^{2}+6 x \geqslant 0$, solving this gives $0 \leqslant x \leqslant 3$. Therefore, $\left(x^{2}+y^{2}+2 x\right)_{\text {max }}=-(3-4)^{2}+16=15$. Hence, the answer is (B).