Four, (20 points) If the hyperbola $y^{2}-x^{2}=1$ and $\frac{x y-x-y+1}{x^{2}-3 x+2}=k$ have a unique common point, find all possible values of $k$.

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$$
\text { IV. From } \begin{aligned}
k & =\frac{x y-x-y+1}{x^{2}-3 x+2} \\
& =\frac{(x-1)(y-1)}{(x-1)(x-2)}
\end{aligned}
$$

we know that when $x \neq 1$, we have
$$
y-1=k(x-2) .
$$

Substituting into the hyperbola equation, we get
$$
\begin{array}{l}
\left(1-k^{2}\right) x^{2}-2 k(1-2 k) x-4 k(k-1) \\
=0 .
\end{array}
$$

There are three cases for the two curves to have a unique common point:
(1) The quadratic term in equation (2) is 0, giving $1-k^{2}$ $=0$, so $k_{1}=1$, with the corresponding common point being $(0,-1) ; k_{2}=$ -1, with the corresponding common point being $\left(\frac{4}{3}, \frac{5}{3}\right)$.
(2) The quadratic equation (2) has equal roots, with the discriminant being 0.
$$
\begin{aligned}
0 & =\Delta=4 k^{2}(1-k)^{2}+16 k(k-1)\left(1-k^{2}\right) \\
& =4 k(5 k-4) .
\end{aligned}
$$

This gives $k_{3}=0$, with the corresponding common point being $(0,1) ; k_{4}=\frac{4}{5}$, with the corresponding common point being $\left(-\frac{4}{3},-\frac{5}{3}\right)$.
(3) The quadratic equation (2) has two distinct real roots, but one of the common points has the x-coordinate $x=1$.
Substituting $x=1$ into (2), we get
$$
\left(1-k^{2}\right)-2 k(1-2 k)-4 k(k-1)=0 \text {, }
$$

which simplifies to $k^{2}-2 k-1=0$.
Solving this, we get $k_{5}=1+\sqrt{2}$, with the corresponding common point being (2 $\sqrt{2}$, $3) ; k_{6}=1-\sqrt{2}$, with the corresponding common point being $(-2 \sqrt{2}$, 3).
In summary, $k$ has 6 values:
$$
1,-1,0, \frac{4}{5}, 1+\sqrt{2}, 1-\sqrt{2} \text {. }
$$