For any given irrational numbers $a, b$ and real number $r \gt 0$, show that there are at most two rational points (points whose coordinates are both rational numbers) on the circle $(x - a)^2 + (y - b)^2 = r^2$.
It is always suggested that askers should post their try but I literally have no idea how to approach this problem. I tried parametrization but it doesn't work.
On
Given three points $A=(x_1,y_1), B=(x_2,y_2), C=(x_3,y_3)$ all co-ordinates rational, the 3 line segments joining them have equations with rational coefficients and rational slope. The midpoints P,Q,R of these 3 line segments also have rational coefficients. Now the circum centre of triangle $ABC$ is the intersection of lines passing through P, Q,R and having slopes negative reciprocal (hence rational) of slopes of AB, BC, CA.
SO the intersection point is rational too.
$(a, b)$ is the center of that circle. If there are three rational points on this circle, you can find their circumcenter by finding the intersection of the perpendicular bisectors, which can be done by solving a system of linear equations with rational coefficients, so $a$ and $b$ must also be rational.