Question

Let $C$ be any circle with centre $(0,\sqrt{2})$. Prove that at the most two rational points can be there on $C$. (A rational point is a point both of whose co-ordinates are rational numbers).

Answer 1

Equation of any circle with centre at $(0,\sqrt{2})$ is given by

$(x-0{)}^2+(y-\sqrt{2})=r^2$

$x^2+y^2-2\sqrt{2}=r^2-2$ ........ $\left(1\right)$

where $r>0$

Let $p\left(x_1{y}_1\right)Q\left(x_2{y}_2\right)R\left(x_3{y}_3\right)$ be three destinct rational points on $\left(1\right)$ since a straight line meet a circle in at most two point either $y_1\ne y_2$

As $P,Q,R$ lie on $\left(1\right)$

$x_1^2+y_1^2-2\sqrt{2}{y}_1=r^2-2$ ....... $\left(2\right)$

$x_2^2+y_2^2-2\sqrt{2}{y}_2=r^2-2$ ......... $\left(3\right)$

$x_3^2+y_3^2-2\sqrt{2}{y}_3=r^2-2$ ....... $\left(4\right)$

$a_1-\sqrt{2}{b}_1=0$ ......... $\left(5\right)$

$a_2-\sqrt{2}{b}_2=0$ ........ $\left(6\right)$

where $a_1=x_2^2+y_2^2-x_1^2-y_1^2$

$a_2=x_3^2+y_3^2-x_1^2-y_1^2$

$b_1=y_2-y_1$

$b_2=y_3-y_1$

Note that $a_1,a_2,a_3$ are rational no. Also either $b_1\ne 0$ or $b_2\ne 0$

Suppose $b_1\ne 0$ then $\sqrt{2}=a_1/b_1$

$\Rightarrow \sqrt{2}$ is a rational no.

This is a contradiction.