Example 6 (1997 Shanghai Competition Problem) On the hyperbola $x y=1$, the point with the abscissa $\frac{n}{n+1}$ is $A_{n}$, and the point with the abscissa $\frac{n+1}{n}$ is $B_{n}\left(n \in \mathbf{N}_{+}\right)$, and the point with coordinates $(1,1)$ is denoted as $M$. Also, $P_{n}\left(x_{n}, y_{n}\right)$ is the circumcenter of $\triangle A_{n} B_{n} M$. When $n \rightarrow+\infty$, find the coordinates $(a, b)$ of the limit point of $P_{n}$, where $a=\lim _{n \rightarrow+\infty} x_{n}, b=\lim _{n \rightarrow+\infty} y_{n}$.
Solving, we get the coordinates of $A_{n}, B_{n}$ as $A_{n}\left(\frac{n}{n+1}, \frac{n+1}{n}\right), B_{n}\left(\frac{n+1}{n}, \frac{n}{n+1}\right)$. Also, $M(1,1)$, so $\left|M A_{n}\right|=\left|M B_{n}\right|=\sqrt{\left(\frac{1}{n+1}\right)^{2}+\left(\frac{1}{n}\right)^{2}}$, thus $\triangle M A_{n} B_{n}$ is an isosceles triangle with the slope of the base $K_{A_{n} B_{n}}=-1$.

Since $M$ lies on the line $y=x$, the equation of the perpendicular bisector of the base is $y=x$, from which we know $x_{n}=y_{n}$. The midpoint of $A_{n} M$ is $E\left(\frac{2 n+1}{2(n+1)}, \frac{2 n+1}{2 n}\right)$, and $K_{A_{n} M}=\frac{\frac{n+1}{n}-1}{\frac{n}{n+1}-1}=-\frac{n+1}{n}$. The circumcenter lies on the line passing through $E$ and perpendicular to $A_{n} M$:
$$
y-\frac{2 n+1}{2 n}=\frac{n}{n+1}\left(x-\frac{2 n+1}{2(n+1)}\right) .
$$

Solving this with $y=x$ gives $x_{n}=\frac{(2 n+1)^{2}}{2 n(n+1)}$, so $\lim _{n \rightarrow+\infty} y_{n}=\lim _{n \rightarrow+\infty} x_{n}=2$, hence the coordinates of the limit point are $(a, b)=(2,2)$.