Let $0<a<b$, and draw lines $l$ and $m$ through the fixed points $A(a, 0)$ and $B(b, 0)$ respectively, such that they intersect the parabola $y^{2}=x$ at four distinct points. When these four points are concyclic, find the locus of the intersection point $P$ of lines $l$ and $m$.
Let the equation of $l$ be $y - kx + ka = 0$,
and the equation of $m$ be $y - k'x + k'b = 0$.

Thus, the quadratic curve passing through the four distinct intersection points of $l$, $m$, and $y^2 = x$ should have the equation
$$
\left(y^2 - x\right) + \lambda(y - kx + ka)(y - k'x + k'b) = 0,
$$

which simplifies to
$$
(1 + \lambda) y^2 - \lambda(k + k') xy + \lambda k k' x^2 + \lambda(ka + k'b) y
- \left(\lambda k k'(a + b) + 1\right) x + \lambda k k' ab = 0
$$

The necessary and sufficient condition for it to be a circle is
$$
\begin{array}{l}
\left\{\begin{array}{l}
k = -k', \\
1 + \lambda = \lambda k k',
\end{array}\right. \\
\left\{\begin{array}{l}
k = -k', \\
\lambda = -\frac{1}{1 + k^2}.
\end{array}\right.
\end{array}
$$

Therefore, the coordinates of the intersection point $P(x_0, y_0)$ of the two lines
$$
\begin{array}{l}
l: y - kx + ka = 0, \\
m: y - k'x + k'b = 0,
\end{array}
$$

are
$$
\left\{\begin{array}{l}
x_0 = \frac{a + b}{2}, \\
y_0 = \frac{k}{2}(b - a).
\end{array}\right.
$$

Thus, point $P$ lies on the perpendicular bisector of segment $AB$.
$\therefore \quad$ The locus of point $P$ is the line $x = \frac{a + b}{2}$, excluding the three intersection points with $y = 0$ or $y^2 = x$.