B. In the Cartesian coordinate system $x O y$, the number of integer point coordinates $(x, y)$ that satisfy the inequality $x^{2}+y^{2} \leqslant 2 x+2 y$ is ( ).
(A) 10
(B) 9
(C) 7
(D) 5
B. B.

From $x^{2}+y^{2} \leqslant 2 x+2 y$, we get
$$
0 \leqslant(x-1)^{2}+(y-1)^{2} \leqslant 2 \text {. }
$$

Since $x$ and $y$ are both integers, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ ( x - 1 ) ^ { 2 } = 0 , } \\
{ ( y - 1 ) ^ { 2 } = 0 ; }
\end{array} \left\{\begin{array}{l}
(x-1)^{2}=0, \\
(y-1)^{2}=1 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ ( x - 1 ) ^ { 2 } = 1 , } \\
{ ( y - 1 ) ^ { 2 } = 0 ; }
\end{array} \left\{\begin{array}{l}
(x-1)^{2}=1, \\
(y-1)^{2}=1 .
\end{array}\right.\right.
\end{array}
$$

Solving, we get $\left\{\begin{array}{l}x=1, \\ y=1 ;\end{array}\left\{\begin{array}{l}x=1, \\ y=2 ;\end{array}\left\{\begin{array}{l}x=1, \\ y=0 ;\end{array}\right.\right.\right.$
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = 0 , } \\
{ y = 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x = 0 , } \\
{ y = 0 ; }
\end{array} \left\{\begin{array}{l}
x=0, \\
y=2 ;
\end{array}\right.\right.\right. \\
\left\{\begin{array}{l}
x=2, \\
y=1 ;
\end{array},\left\{\begin{array}{l}
x=2, \\
y=0 ;
\end{array},\left\{\begin{array}{l}
x=2, \\
y=2 .
\end{array}\right.\right.\right.
\end{array}
$$