$37 \cdot 4$ In the Cartesian coordinate system, a point whose both coordinates are rational numbers is called a rational point. If $a$ is an irrational number, then among all lines passing through the point $(a, 0)$,
(A) there are infinitely many lines, each of which contains at least two rational points.
(B) there are exactly $n(2 \leqslant n<+\infty)$ lines, each of which contains at least two rational points:
(C) there is exactly one line that passes through at least two rational points.
(D) each line passes through at most one rational point.
(China High School Mathematics League, 1987)
[Solution] The equation of a line passing through the point $(a, 0)$ is $x=a$ and $y=k(x-a)$, where $-\infty<k<+\infty$. There are no rational points on the line $x=a$.
Now consider the line $y=k(x-a)$.
When $k=0$, the line $y=0$ has infinitely many rational points, thus negating conclusion $(D)$;
When $k \neq 0$, if there are two rational points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ on this line, then from
$$
y_{1}=k\left(x_{1}-a\right) \text {, and } y_{2}=k\left(x_{2}-a\right) \text {, }
$$

we get $y_{2}-y_{1}=k\left(x_{2}-x_{1}\right)$, which implies that $k$ must be a rational number.
However, when $k$ is a rational number and $k \neq 0$, there are no rational points on the line $y=k(x-a)$ (note that if $x, y, k$ are all rational numbers, $a$ must also be a rational number, which contradicts the given condition), thus negating conclusions (A) and (B).
Therefore, the correct choice is $(C)$.