II. 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 $B_{n}(n \in N)$. The point with coordinates $(1,1)$ is denoted as $M, P_{n}\left(x_{n}, y_{n}\right)$ is the circumcenter of $\triangle A_{n} B_{n} M$. Find the coordinates $(a, b)$ of the limit point of $P_{n}$ as $n \rightarrow \infty$, where $a=\lim _{n \rightarrow \infty} x_{n}, b=\lim _{n \rightarrow \infty} y_{n}$.
II. It is easy to get $A_{n}\left(\frac{n}{n+1}, \frac{n+1}{n}\right), B_{n}\left(\frac{n+1}{n}, \frac{n}{n+1}\right)$.
$\therefore\left|A_{n} M\right|=\left|B_{n} M\right|$, and $k_{A_{n} B_{n}}=-1$.
Thus, $\triangle M A_{n} B_{n}$ is an isosceles triangle with $A_{n} B_{n}$ as the base, and the slope of the altitude is -1. Since $M$ lies on the line $y=x$, the equation of the median of the base is $y=x$. Therefore, $x_{n}=y_{n}$.
The midpoint of $A_{n} M$ is $E\left(\frac{2 n+1}{2(n+1)}, \frac{2 n+1}{2 n}\right)$,
$k_{1_{n} n}-\frac{\frac{n+1}{n}-1}{-\frac{n}{n+1}-1}=-\frac{n+1}{n}$.
Thus, the circumcenter $P_{n}\left(x_{n}, x_{n}\right)$ lies on the line $y-\frac{2 n+1}{2 n}=\frac{n}{n+1}$
- $\left(x-\frac{2 n+1}{2(n+1)}\right)$, from which we get $x_{n}=\frac{(2 n+1)^{2}}{2 n(n+1)}$.
$$
\therefore \lim _{n \rightarrow \infty} x_{n}=\lim _{n \rightarrow \infty} y_{n}=2 \text {. }
$$

Thus, the coordinates of the limit point are $(2,2)$.