Question

Prove that the maximum number of points with rational coordinates on a circle whose center is $\left(\sqrt{3},0\right)$ is two

Answer 1

Given centre is $(\sqrt{3},0)$ and equation of circle is $(x-h{)}^2+(y-k{)}^2=r^2$

$\therefore$ Equation of circle is $(x-\sqrt{3}{)}^2+(y-0{)}^2=r^2$

$(x-\sqrt{3}{)}^2=r^2-y^2$

$\Rightarrow x=\sqrt{3}\pm \sqrt{r^2-y^2}$

Coordinate with rational if x is rational then $x=\sqrt{3}+\sqrt{r^2-y^2}$ [always irrational]

$x=\sqrt{3}-\sqrt{r^2-y^2}$, when $\sqrt{r^2-y^2}=\sqrt{3}$

We get $x=0$

$\Rightarrow \sqrt{r^2-y^2}=\sqrt{3}$

$\Rightarrow r^2-y^2=3$

This is possible only on two points at $r=2$, $y=1$

and $r=2$, $y=-1$

$\therefore$ The rational co-ordinates are $(0,1)$ and $(0,-1)$.