The rectangular coordinates of the points on a line are given by the following expressions:
(1) $\quad x=\frac{1-u^{2}-v^{2}}{(1-u)^{2}+v^{2}}$

$$
y=\frac{2 v}{(1-u)^{2}+v^{2}}
$$

where $u, v$ are parameters and $v / u=m, m$ is a non-zero constant. Find the equation of the line in terms of the relationship between $x$, $y$, and $m$.
I. solution. Since $m \neq 0$, it follows that $v \neq 0$, and thus $y \neq 0$ always holds, and also $u \neq 0$. The expressions are always defined because their common denominator is not less than $v^2$, which is positive. From (1),

$$
x+1=\frac{2(1-u)}{(1-u)^{2}+v^{2}}
$$

dividing this by (2),

$$
\frac{x+1}{y}=\frac{1-u}{v}=\frac{1-u}{m u}
$$

and from this, we always have

$$
u=\frac{y}{y+m(x+1)}, \quad v=\frac{m y}{y+m(x+1)}
$$

indeed, the denominator does not vanish because, based on (2) and (3),

$$
y+m(x+1)=\frac{2 m}{(1-u)^{2}+v^{2}} \neq 0
$$

Based on these, we can easily obtain the desired form of the relationship between $x, y$, and $m$. From (4),

$$
1-u=\frac{m(x+1)}{y+m(x+1)}
$$

and thus from the rearranged form of (2),

$$
(1-u)^{2}+v^{2}=\frac{2 v}{y}
$$

we get:

$$
\begin{gathered}
\frac{m^{2}(x+1)^{2}}{[y+m(x+1)]^{2}}+\frac{m^{2} y^{2}}{[y+m(x+1)]^{2}}=\frac{2 m}{y+m(x+1)} \\
(x+1)^{2}+y^{2}=\frac{2}{m}[y+m(x+1)] \\
x^{2}+\left(y-\frac{1}{m}\right)^{2}=1+\frac{1}{m^{2}}
\end{gathered}
$$

This is the equation of a circle, whose center $K$ and radius $r$ are:

$$
K\left(0, \frac{1}{m}\right), \quad r=\sqrt{1+\frac{1}{m^{2}}}
$$

However, we have excluded the value $y=0$, which corresponds to $x = \pm 1$ according to (5). Therefore, the desired curve is the circle described above, excluding the points $(1,0)$ and $(-1,0)$ on the $x$-axis (the upper and lower arcs of the circle).

Remarks. 1. We can also obtain (5) with the following reasoning. For every point on the desired curve, there is at least one pair of values for the parameters $u, v$ that satisfy the given conditions, and considering the relationship $v = m u$ and the constancy of $m$, there is at least one value for $u$. Each coordinate of this point gives 2 values, because from (1) and (2), rearranging for $u$ (which is an equivalent transformation according to the assumptions):

$$
\begin{gathered}
\left(1+m^{2}\right)(x+1) u^{2}-2 x u+(x-1)=0 \\
\quad\left(1+m^{2}\right) y u^{2}-2(m+y) u+y=0
\end{gathered}
$$

This means that (6) and (7) have a common root.

It is expected that in such a case, there is a relationship between the coefficients of the two equations, since for arbitrary coefficients, two quadratic equations do not generally have the same roots. We seek this relationship for the pair of equations,

$$
\begin{gathered}
a_{1} z^{2}+b_{1} z+c_{1}=0 \\
a_{2} z^{2}+b_{2} z+c_{2}=0
\end{gathered}
$$

Let their common root be $z_{1}$, then we have

$$
\begin{aligned}
& a_{1} z_{1}^{2}+b_{1} z_{1}+c_{1}=0 \\
& a_{2} z_{1}^{2}+b_{2} z_{1}+c_{2}=0
\end{aligned}
$$

First, eliminate $z_{1}^{2}$ and then $z_{1}$ (by multiplying (8)