2. Let $a$ be a real number, and the two parabolas
$$
y=x^{2}+x+a \text { and } x=4 y^{2}+3 y+a
$$

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.
2. (1) Given the subtraction and rearrangement of two equations, we have
$$
\begin{array}{l}
(x-2 y)(x+2 y+2)=0 \\
\Rightarrow x-2 y=0 \text { or } x+2 y+2=0 .
\end{array}
$$

From this, we know that the two parabolas in the problem have four intersection points if and only if the system of equations
$$
\left\{\begin{array} { l } 
{ y = x ^ { 2 } + x + a , } \\
{ x - 2 y = 0 }
\end{array} \text { and } \left\{\begin{array}{l}
y=x^{2}+x+a, \\
x+2 y+2=0
\end{array}\right.\right.
$$

has four distinct real solutions. This is equivalent to each system of equations having two distinct real solutions and no common solutions.

From the first system of equations, we get $2 x^{2}+x+2 a=0$, and its discriminant is greater than 0, i.e.,
$$
1-16 a>0 \Rightarrow a<\frac{1}{16} \text {. }
$$

From the second system of equations, we get $2 x^{2}+5 x+2 a+2=0$, and its discriminant is greater than 0, i.e.,
$$
25-16(2 a+2)>0 \Rightarrow a<-\frac{7}{16} \text {. }
$$

Therefore, each system of equations has two distinct real solutions if and only if $a<-\frac{7}{16}$.
Note that the two systems of equations have a common solution if and only if
$$
\left\{\begin{array} { l } 
{ y = x ^ { 2 } + x + a , } \\
{ x - 2 y = 0 , } \\
{ x + 2 y + 2 = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=-1, \\
y=-\frac{1}{2}, \\
a=-\frac{1}{2} .
\end{array}\right.\right.
$$

Therefore, $a \in\left(-\infty,-\frac{1}{2}\right) \cup\left(-\frac{1}{2},-\frac{7}{16}\right)$.
(2) From $y=x^{2}+x+a$, we get
$4 y=4 x^{2}+4 x+4 a$.
Adding and rearranging with $x=4 y^{2}+3 y+a$, we get
$$
\left(x+\frac{3}{8}\right)^{2}+\left(y-\frac{1}{8}\right)^{2}=-\frac{5 a}{4}+\frac{5}{32} \text {. }
$$

This indicates that the four intersection points lie on a circle, with the center coordinates being $\left(-\frac{3}{8}, \frac{1}{8}\right)$.