I'd think that if a rational point has both x,y rational, an irrational point would be
a point that is not ratinal, so has at least one of x,y irratinal?
write down the equations of the circle for two points
[tex](X - x_1)^2 + (Y - y_1)^2 = R^2[/tex]
[tex](X - x_2)^2 + (Y - y_2)^2 = R^2[/tex]
You can eliminate R from them, and write them so X and Y become separated
[tex]2 X (x_2 - x_1) + 2 Y (y_2 - y1) = x_2^2 + y_2^2 - x_1^2 - y_1^2[/tex]
Note that all numbers, except for X and Y are rational. With 2 points it's still
possible to have X or Y irrational. Now add a third point
[tex](X - x_3)^2 + (Y - y_3)^2 = R^2[/tex]
and combine this equation with the one for the first point producing
[tex]2 X (x_3 - x_1) + 2 Y (y_3 - y1) = x_3^2 + y_3^2 - x_1^2 - y_1^2[/tex]
we now get 2 simultaneous linear equations for X and Y, and it is possible to prove that X and Y cannot be rational unless 2 of the rational points on the circle are identical, or the rational points on the circle lie on a straight line.