JEE Advance - Mathematics (1997 - No. 18)
Let C be any circle with centre $$\,\left( {0\, , \sqrt {2} } \right)$$. Prove that at the most two rational points can to there on C. (A rational point is a point both of whose coordinates are rational numbers.)
The statement is false; there exist circles centered at (0, √2) with three or more rational points.
If (x, y) is a rational point on the circle, then (x - 0)^2 + (y - √2)^2 = r^2 for some radius r.
Expanding the equation gives x^2 + y^2 - 2√2y + 2 = r^2, which implies 2√2y = x^2 + y^2 + 2 - r^2.
If three rational points exist on the circle, we can create a contradiction by showing that √2 must be rational.
If two distinct rational points lie on the circle, the slope of the perpendicular bisector of the line segment connecting them must be rational.