$37 \cdot 6$ points $(x, y)$ where both $x, y$ are rational numbers are called rational points. In a circle with radius $r$ and center $(a, b)$, if $a$ is a rational number and $b$ is an irrational number, the number of rational points on this circle is
(A) at most one.
(B) at most two.
(C) at most three.
(D) infinitely many.
(1st "Hope Cup" National Mathematics Invitational Competition, 1990)
[Solution] Let the equation of the circle be $(x-a)^{2}+(y-b)^{2}=r^{2}$. Now consider two points on the circle $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$. Substituting these points into the equation and subtracting, then rearranging, we get $\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}-2 a\right)+\left(y_{1}-y_{2}\right)\left(y_{1}+y_{2}-2 b\right)=0$. If both points are rational points, then $y_{1}+y_{2}-2 b \neq 0$. However, when $y_{1}=y_{2}$, if $x_{1}+x_{2}=2 a$, the equation $\circledast$ still holds. This means that the circle can have two rational points.
If the circle has another rational point $\left(x_{3}, y_{3}\right)$, then it must be that $y_{1}=y_{2}=y_{3}$, which is impossible.
Therefore, the answer is $(B)$.