Example 8 Let $0<a<b$, 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: $y-0=k(x-a)$, and the equation of $m$ be: $y-0=k'(x-b)$. Then the quadratic curve passing through the four distinct intersection points of $l$, $m$, and $y^2 = x$ has 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: $k = -k'$, $1 + \lambda = \lambda k k'$, i.e., 
$$\left\{\begin{array}{l}k = -k' \\ \lambda = -\frac{1}{1 + k^2}\end{array}\right..$$
Thus, the coordinates of the intersection point $P(x_0, y_0)$ of the two moving lines $l$ and $m$ are 
$$\left\{\begin{array}{l}x_0 = \frac{a + b}{2}, \\ y_0 = \frac{k}{2}(a - b).\end{array}\right.$$
This indicates that $P$ lies on the perpendicular bisector of segment $AB$, and $y_0 \neq 0$, $y_0^2 \neq x_0$ (otherwise, $l$, $m$, and the parabola would not have four intersection points). Therefore, the trajectory of point $P$ is the line $x = \frac{a + b}{2}$, excluding the three intersection points with $y = 0$ and $y^2 = x$.