As shown in Figure 8, let the intersection point of the circle and the parabola be $p\left(x_{0}, y_{0}\right)$, and the center of the circle be point $\mathrm{C}$.

From the problem, the tangent line to the parabola at point $\mathrm{P}$ must pass through the center $\mathrm{C}$. The equation of the tangent line $\mathrm{PC}$ is
$$y_{0} y = p\left(x + x_{0}\right).$$

Since point $\mathrm{C}$ is the intersection of the tangent line $\mathrm{PC}$ and the $x$-axis,
$$\left\{\begin{array}{l} y_{0} y = p\left(x + x_{0}\right), \\ y = 0 \end{array}\right.$$

Solving this, we get the coordinates of the center $\mathrm{C}$ as $\left(-x_{0}, 0\right)$.
Therefore, the equation of the circle is
$$\left(x + x_{0}\right)^{2} + y^{2} = 1.$$

Next, we find the value of $x_{0}$:
Since $\mathrm{P}\left(x_{0}, y_{0}\right)$ is the intersection point of the circle and the parabola,
$$\left\{\begin{array}{l} \left(x + x_{0}\right)^{2} + y^{2} = 1, \\ y^{2} = 2 x \end{array}\right.$$

By the meaning of the solution, we have
$$\left\{\begin{array}{l} \left(x_{0} + x_{0}\right)^{2} + y_{0}^{2} = 1, \\ y_{0}^{2} = 2 x_{0}. \end{array}\right.$$

Eliminating $y_{0}$, we get
$$\begin{array}{c} 4 x_{0}^{2} + 2 x_{0} - 1 = 0, \\ x_{0} = \frac{-1 \pm \sqrt{5}}{4}. \\ \text{Since } x_{0} > 0, \quad \therefore x_{0} = \frac{-1 + \sqrt{5}}{4}. \end{array}$$

Therefore, when the center of the circle moves to $\left(-\frac{\sqrt{5} - 1}{4}, 0\right)$, the tangents to the circle and the parabola at the intersection point are perpendicular.