12. (16 points) On the Cartesian plane, a point whose both coordinates are rational numbers is called a rational point. Find the smallest positive integer $k$ such that: for every circle that contains $k$ rational points on its circumference, the circle must contain infinitely many rational points on its circumference.
12. First, prove: If a circle's circumference contains 3 rational points, then the circumference must contain infinitely many rational points.

Let $\odot C_{0}$ in the plane have 3 rational points $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$ on its circumference, with the center $C_{0}\left(x_{0}, y_{0}\right)$.

Since the perpendicular bisectors of segments $P_{1} P_{2}$ and $P_{1} P_{3}$ pass through the center $C_{0}$, we have
$$
\left\{\begin{array}{l}
\left(y_{2}-y_{1}\right)\left(y_{0}-\frac{y_{1}+y_{2}}{2}\right)+\left(x_{2}-x_{1}\right)\left(x_{0}-\frac{x_{1}+x_{2}}{2}\right)=0, \\
\left(y_{3}-y_{1}\right)\left(y_{0}-\frac{y_{1}+y_{3}}{2}\right)+\left(x_{3}-x_{1}\right)\left(x_{0}-\frac{x_{1}+x_{3}}{2}\right)=0 .
\end{array}\right.
$$

Since $x_{i} 、 y_{i}(i=1,2,3)$ are all rational numbers, the solution $\left(x_{0}, y_{0}\right)$ of the above system of linear equations in $x_{0} 、 y_{0}$ is also a rational number, i.e., $C_{0}$ is a rational point.
Suppose the coordinates of the rational point $P_{n}\left(x_{n}, y_{n}\right)$ are
$$
\begin{array}{l}
\qquad\left\{\begin{array}{l}
x_{n}=x_{0}+a_{n}\left(x_{3}-x_{0}\right)-b_{n}\left(y_{3}-y_{0}\right), \\
y_{n}=y_{0}+b_{n}\left(x_{3}-x_{0}\right)+a_{n}\left(y_{3}-y_{0}\right),
\end{array}\right. \\
\text { where, } a_{n}=\frac{n^{2}-1}{n^{2}+1}, b_{n}=\frac{2 n}{n^{2}+1}(n=4,5, \cdots) .
\end{array}
$$

Then $\left|P_{n} C_{0}\right|^{2}$
$$
\begin{aligned}
= & {\left[a_{n}\left(x_{3}-x_{0}\right)-b_{n}\left(y_{3}-y_{0}\right)\right]^{2}+} \\
& {\left[b_{n}\left(x_{3}-x_{0}\right)+a_{n}\left(y_{3}-y_{0}\right)\right]^{2} } \\
= & \left(a_{n}^{2}+b_{n}^{2}\right)\left[\left(x_{3}-x_{0}\right)^{2}+\left(y_{3}-y_{0}\right)^{2}\right] \\
= & \left(x_{3}-x_{0}\right)^{2}+\left(y_{3}-y_{0}\right)^{2}=\left|P_{3} C_{0}\right|^{2} .
\end{aligned}
$$

Hence, the points $P_{n}(n=4,5, \cdots)$ are all on the circumference of $\odot C_{0}$, i.e., the circumference of $\odot C_{0}$ contains infinitely many rational points.

Second, construct an example of a circle whose circumference contains only two rational points.
$$
C:(x-\sqrt{2})^{2}+(y-\sqrt{2})^{2}=6 .
$$

It is easy to verify that $P_{1}(-1,1)$ and $P_{2}(1,-1)$ are on the circumference of $C$. If the circumference of $C$ contains another rational point $P_{3}\left(x_{3}, y_{3}\right)$ different from $P_{1}$ and $P_{2}$, then
$$
\left(x_{3}-\sqrt{2}\right)^{2}+\left(y_{3}-\sqrt{2}\right)^{2}=6 \text {, }
$$

i.e., $x_{3}^{2}+y_{3}^{2}-2=2 \sqrt{2}\left(x_{3}+y_{3}\right)$.
Since the left side is a rational number and $\sqrt{2}$ is an irrational number, we have $x_{3}+y_{3}=0$. Thus, $x_{3}^{2}=1$. Hence, $x_{3}= \pm 1, y_{3}=$ $\mp 1$. This contradicts the assumption that $P_{3}$ is different from $P_{1}$ and $P_{2}$.
In conclusion, the minimum value of $k$ is 3.