Example 11. The equation of the parabola is $y^{2}=2 x$, and there is a circle with a radius of 1, whose center
moves on the $x$-axis. Ask at what position
does the circle move to
when the tangents at the intersection points
of the circle and the parabola
are perpendicular to each other?
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.