11. Given the circle $O: x^{2}+y^{2}=4$ and the curve $C: y=3|x-t|, A(m, n), B(s, p)\left(m, n, s, p \in \mathrm{N}^{*}\right)$ are two points on the curve $C$, such that the ratio of the distance from any point on circle $O$ to point $A$ and to point $B$ is a constant $k(k>1)$, find the value of $t$.
【Analysis】Answer: $t=\frac{4}{3}$.
Take points $C(2,0), D(-2,0), E(0,2), F(0,-2)$ on the circle, according to the problem, we have:
$$
\left\{\begin{array}{l}
\frac{(2-m)^{2}+n^{2}}{(2-s)^{2}+p^{2}}=\frac{(2+m)^{2}+n^{2}}{(2+s)^{2}+p^{2}}=\frac{m}{s}, \\
\frac{m^{2}+(2-n)^{2}}{s^{2}+(p-2)^{2}}=\frac{m^{2}+(2+n)^{2}}{s^{2}+(2+p)^{2}}=\frac{n}{p},
\end{array}\right.
$$

Thus: $\overrightarrow{O A}=t \overrightarrow{O B}$, so points $A, B, O$ are collinear.
By the property of the Apollonian circle: $O A \cdot O B=\mathrm{R}^{2}=4$, and $O A=\frac{R}{\lambda}, O B=R \lambda$, where $\lambda>1$, then $O A<O B$, so: $O A<2$;

Since: $m^{2}+n^{2}=O A^{2}=\frac{4}{\lambda^{2}}$,
and $m, n \in \mathrm{N}^{*}$, thus: $O A^{2}=\frac{4}{\lambda^{2}} \in \mathrm{N}^{*}$,
(1) If $O A^{2}=\frac{4}{\lambda^{2}}=1$, then $\lambda=2$, at this time: $m^{2}+n^{2}=1$, there must be $m n=0$, because $m, n \in \mathrm{N}^{*}$, which does not meet the problem's requirements;
(2) If $O A^{2}=\frac{4}{\lambda^{2}}=2$, then $\lambda=\sqrt{2}$, at this time: $m^{2}+n^{2}=2$, we get: $m=n=1, s=p=2$, the equation of line $A B$ is $y=x$, then points $A(1,1), B(2,2)$ are on curve $C$, substituting and solving gives: $t=\frac{4}{3}$.
(3) If $O A^{2}=\frac{4}{\lambda^{2}}=3$, at this time: $m^{2}+n^{2}=3$, there is no positive integer solution, which does not meet the problem's requirements.

In summary: $t=\frac{4}{3}$.