1. (1) If the parabola $y=x^{2}-a$ intersects the hyperbola $y=\frac{1}{x}$ at three distinct points, find the range of real values for $a$.
(2) In (1), when $a$ takes all possible values, find the locus of the circumcenter of the triangle formed by the three intersection points.
1. (1) As shown in Figure 6, we study the case where a parabola is tangent to a hyperbola.
Let $\alpha$ be the x-coordinate of the tangent point. Then
$$
2 \alpha = -\frac{1}{\alpha^{2}} \Rightarrow \alpha = -\frac{1}{\sqrt[3]{2}}
$$
$\Rightarrow$ The y-coordinate of the tangent point is $\frac{1}{\alpha} = -\sqrt[3]{2}$.
Substituting into the parabola equation gives $a = \frac{3 \sqrt[3]{2}}{2}$.
Clearly, when $a > \frac{3 \sqrt[3]{2}}{2}$, they intersect at exactly three points.
(2) Each intersection point satisfies $y = x^{2} - a, x y = 1$.
Thus, $y^{2} = x^{2} y - a y = x - a y$.
From $y = x^{2} - a$
$\Rightarrow y - y^{2} = x^{2} - a - x + a y$
$\Rightarrow \left(x - \frac{1}{2}\right)^{2} + \left(y + \frac{a - 1}{2}\right)^{2} = \frac{(a + 1)^{2} + 1}{4}$.
Therefore, the circumcenter of the triangle formed by these three points is $\left(\frac{1}{2}, -\frac{a - 1}{2}\right)$.
Hence, the locus of the circumcenter is a ray
$$
x = \frac{1}{2} \text{, and } y < -\frac{3 \sqrt[3]{2}}{4} + \frac{1}{2} \text{. }
$$