21. When the value range of $a$ is $(\quad)$, the curve $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ intersect at exactly three points in the coordinate plane.
(A) $a=\frac{1}{4}$
(B) $\frac{1}{4}<a<\frac{1}{2}$
(C) $a>\frac{1}{4}$
(D) $a=\frac{1}{2}$
(E) $a>\frac{1}{2}$
21. E.

From the problem, combining the equations of the two curves, we get
$$
\begin{array}{l}
x^{2}+\left(x^{2}-a^{2}\right)=a^{2} \\
\Rightarrow x^{4}-(2 a-1) x^{2}=0 \\
\Rightarrow\left(x^{2}-(2 a-1)\right) x^{2}=0 \\
\Rightarrow x^{2}-(2 a-1)=0 .
\end{array}
$$

Since the curves $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ intersect at exactly three points in the coordinate plane, the equation (2) has two distinct real roots, i.e., $\square$
$$
x= \pm \sqrt{2 a-1} \Rightarrow a>\frac{1}{2} .
$$