Example 6 Given $0<a<b$, 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 $m$.
(1993, National High School Mathematics League Competition)
Let $l: y=k_{1}(x-a), m: y=k_{2}(x-b)$. If the four intersection points are concyclic, then the circle can be written as
$$
\begin{array}{l}
\left(k_{1} x-y-k_{1} a\right)\left(k_{2} x-y-k_{2} b\right)+ \\
\lambda\left(y^{2}-x\right)=0,
\end{array}
$$

In this equation, the $x y$ term must be 0, hence we get $k_{1}=-k_{2}$.
Then $l: y=k_{1}(x-a), m: y=-k_{1}(x-b)$.
Eliminating $k_{1}$ from the two equations, we get
$$
2 x-(a+b)=0(y \neq 0) \text {, }
$$

This is the required equation of the locus.