Confusing Conics & Circles! Wow!!

The general equation of circle satisfying the above condition can be written as

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

where r is the radius of circle,expanding it we get

${ x }^{ 2 }+{ y }^{ 2 }+2\quad =\quad { r }^{ 2 }+2\sqrt { 2 } y$

Since x,y are rational ,Right hand side must also be rational

$Case:1$ - ${ r }^{ 2 }$ is rational

Then only y=0 will satisfy given condition and give at most 2 points.

$Case:2$ - ${ r }^{ 2 }$ is irrational

Sum of 2 irrational numbers is always irrational unless ${ r }^{ 2 }$ is of form $P-2\sqrt { 2 } Q$ (where P and Q are rational) but even then we get rational number on right side only for y=Q and get at most two points.

Let there be three rational points on circle C. Then there Circumcentre should come rational because circumcentre is intersection of perpendicular bisectors .You can figure it out mathematically too. But given circumcentre is irrational so 3 rational points are not possible.

Now considering 4 or more rational points Here take any three points and the previous case will arise.

Hence at most 2 rational points are possible.

please post contradictions and comments and questions.

try this question and have fun and please reshare it if you like it.

Let us assume that there are $3$ rational points on the circle having its equation: $x^{2}+y^{2}+2gx+2fy+c=0$ . Since, three rational points lie on the circle, we will get three linear equations in $g,f$ and $c$ with rational coefficients. But $f=\sqrt{2}$ . Thus, we arrive at a contradiction. Hence, there can be at most 2 rational points on $C$ .