11. (15 points) Given $\odot O: x^{2}+y^{2}=4$ and the curve $C: y=3|x-t|, A(m, n) 、 B(s, p)(m 、 n 、 s 、 p \in$ $\left.\mathbf{Z}_{+}\right)$ are two points on the curve $C$, such that the ratio of the distance from any point on $\odot O$ to point $A$ and to point $B$ is a constant $k(k>1)$. Find the value of $t$.
11. Let $P(x, y)$ be any point on $\odot O$.
From the problem, we have
$$
\begin{array}{l}
\frac{|P A|}{|P B|}=k \Rightarrow P A^{2}=k^{2} P B^{2} \\
\Rightarrow(x-m)^{2}+(y-n)^{2}=k^{2}\left((x-s)^{2}+(y-p)^{2}\right) \\
\Rightarrow x^{2}+y^{2}-\frac{2\left(k^{2} s-m\right)}{k^{2}-1} x-\frac{2\left(k^{2} p-n\right)}{k^{2}-1} y \\
\quad=\frac{\left(m^{2}+n^{2}\right)-k^{2}\left(s^{2}+p^{2}\right)}{k^{2}-1} .
\end{array}
$$

Thus, the locus of point $P$ is a circle.
Since $P(x, y)$ is any point on the circle, this circle must be the same as $\odot O$.
$$
\begin{array}{l}
\text { Hence } \frac{-2\left(k^{2} s-m\right)}{k^{2}-1}=0, \frac{-2\left(k^{2} p-n\right)}{k^{2}-1}=0, \\
\frac{\left(m^{2}+n^{2}\right)-k^{2}\left(s^{2}+p^{2}\right)}{k^{2}-1}=4 .
\end{array}
$$

Simplifying, we get
$$
\begin{array}{l}
k^{2} s-m=0, k^{2} p-n=0 \\
\Rightarrow \frac{\left(m^{2}+n^{2}\right)-k^{2}\left(s^{2}+p^{2}\right)}{k^{2}-1} \\
=\frac{\left(k^{4} s^{2}+k^{4} p^{2}\right)-k^{2}\left(s^{2}+p^{2}\right)}{k^{2}-1} \\
=k^{2}\left(s^{2}+p^{2}\right)=4 .
\end{array}
$$

Since $s, p \in \mathbf{Z}_{+}$, we have $s^{2} \geqslant 1, p^{2} \geqslant 1$.
Thus, $k^{2}=\frac{4}{s^{2}+p^{2}} \leqslant 2$.
Also, $k>1$, so $s=p=1, k^{2}=2, m=n=2$.
Therefore, $A(2,2), B(1,1)$.
Substituting into $y=3|x-t|$, we get $t=\frac{4}{3}$.