14. Let $a$ be a real number, and the two parabolas $y=x^{2}+x+a$ and $x=4 y^{2}+3 y+a$ have four intersection points.
(1) Find the range of values for $a$;
(2) Prove that these four intersection points are concyclic, and find the coordinates of the center of the circle.
(1) Solving $\left\{\begin{array}{l}y=x^{2}+x+a, \\ x=4 y^{2}+3 y+a\end{array}\right.$
$\Rightarrow x^{2}-4 y^{2}+2 x-4 y=0$
$\Rightarrow(x-2 y)(x+2 y+2)=0$, thus the two parabolas have
four intersection points if and only if $y=x^{2}+x+a$ intersects with the two lines
$x-2 y=0$ and $x+2 y+2=0$ at two distinct points each.
Solving $\left\{\begin{array}{l}x-2 y=0, \\ y=x^{2}+x+a\end{array} \Rightarrow 2 x^{2}+x+2 a=0 \Rightarrow \Delta_{1}=1-16 a>0 \Rightarrow a<\frac{1}{16}\right.$,
and when the intersection points coincide, $\left\{\begin{array}{l}x-2 y=0, \\ x+2 y+2=0,\end{array} \Rightarrow\left\{\begin{array}{l}x=-1, \\ y=-\frac{1}{2},\end{array} \Rightarrow a=-\frac{1}{2}\right.\right.$.
In summary, the range of $a$ is $\left(-\infty,-\frac{1}{2}\right) \cup\left(-\frac{1}{2},-\frac{7}{16}\right)$.
(2) Solving $\left\{\begin{array}{l}4 y=4 x^{2}+4 x+4 a, \\ x=4 y^{2}+3 y+a\end{array} \Rightarrow 4 x^{2}+4 y^{2}+3 x-y+5 a=0\right.$
$\Rightarrow\left(x+\frac{3}{8}\right)^{2}+\left(y-\frac{1}{8}\right)^{2}=\frac{5(1-8 a)}{32}$. This means the four intersection points lie on a circle, with the center coordinates at $\left(-\frac{3}{8}, \frac{1}{8}\right)$.