Given the equation of a circle as $x^{2}+y^{2}=4$. Try to find two points $A(s, t), B(m, n)$ on the coordinate plane, such that the following two conditions are satisfied:
(1) The ratio of the distance from any point on the circle to point $A$ and to point $B$ is a constant $k ;$.
(2) $s>m, t>n$, and $m, n$ are both natural numbers.
Let any point on the circle be $P(x, y)$, then
$$
\frac{\sqrt{(s-x)^{2}+(1-y)^{2}}}{\sqrt{(m-x)^{2}+(n-y)^{2}}}=k .
$$

Taking $P_{1}(2,0), P_{2}(-2,0)$, we get
$$
\left.\begin{array}{l}
\frac{(s-2)^{2}+t^{2}}{(m-2)^{2}+n^{2}}=k^{2} \\
\frac{(s+2)^{2}+t^{2}}{(m+2)^{2}+n^{2}}=k^{2}
\end{array}\right\} \Rightarrow S=k^{2} m ;
$$

Taking $P_{3}(0,2), P_{1}(0,-2)$, we get
$$
t=k^{2} n \text {. }
$$

Substituting (2) and (3) into (1), and taking $x=y=\sqrt{2}$, we have
$$
\frac{\left(k^{2} m-\sqrt{2}\right)^{2}+\left(k^{2} n-\sqrt{2}\right)^{2}}{(m-\sqrt{2})^{2}+(n-\sqrt{2})^{2}}=k^{2},
$$

Simplifying to
$$
k^{2}\left(m^{2}+n^{2}\right)-4=0 . \therefore m=n=1, k^{2}=2 \text {. }
$$

The required points are $(2,2),(1.1)$.