* 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 $B_{n}(n \in \mathbf{N})$. 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 limit point coordinates $(a, b)$ of $P_{n}$, where $a=\lim _{n \rightarrow \infty} x_{n}, b=\lim _{n \rightarrow \infty} y_{n}$.
It is easy to obtain the coordinates of $A_{n}$ and $B_{n}$: $A_{n}\left(\frac{n}{n+1}, \frac{n+1}{n}\right), B_{n}\left(\frac{n+1}{n}, \frac{n}{n+1}\right)$. Therefore, $\left|A_{n} M\right|=\left|B_{n} M\right|$, and the slope of the line $A_{n} B_{n}$ is $k_{A_{n} B_{n}}=-1$. Hence, $\triangle M A_{n} B_{n}$ is an isosceles triangle with base $A_{n} B_{n}$, and the slope of the base is -1.

Since $M$ lies on the line $y=x$, the equation of the perpendicular bisector of the base is $=x$. This implies $x_{n}=y_{n}$. The midpoint of $A_{n} M$ is $E\left(\frac{2 n+1}{2(n+1)}, \frac{2 n+1}{2 n}\right)$, and $k_{A_{n} M}=\frac{\frac{n+1}{n}-1}{\frac{n}{n+1}-1}=-\frac{n+1}{n}$. Therefore, 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 find $x_{n}=\frac{(2 n+1)^{2}}{2 n(n+1)}$. Thus, $\lim _{n \rightarrow \infty} x_{n}=\lim _{n \rightarrow \infty} y_{n}=2$. Therefore, the coordinates of the limit point are $(2,2)$.