121 The solutions to the inequality $x(x-1) \leqslant y(1-y)$ can all make $x^{2}+y^{2} \leqslant k$ hold, the minimum value of $k$ is $(\quad)$
A. 0
B. 1
C. 2
D. Greater than 2
121 C. The inequality $x(x-1) \leqslant y(1-y)$ is equivalent to $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2} \cdot x^{2}+y^{2}=$ $(\sqrt{k})^{2}$, which represents a circle with radius $\sqrt{k}$. When this circle is internally tangent to $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$, $k$ takes its minimum value.