Let `C` be any circle with centre `(0,sqrt(2))dot` Prove that at most two rational points can be there on `Cdot` (A rational point is a point both of whose coordinates are rational numbers)

Explanation

Equations of any circle C with centre at `(0, sqrt2)` is given by
`(x-0)^(2) + (y-sqrt2) = r^(2)`
or `x^(2)+y^(2)-2sqrt2 y + 2 = r^(2) " "...(i)`
where, ` r gt 0`
Let `(x_(1), y_(1)), (x_(2), y_(2)), (x_(3),y_(3))` be three distinct rational points on circle. Since a straight line parallel to X-axis meets a circle in at most tow points, either `y_(1), y_(2) or y_(1), y_(3)`.
On putting these in Eq (i), we get
`x_(1)^(2)+y_(1)^(2) -2sqrt2 y_(1) = r^(2) - 2" "...(ii)`
`x_(2)^(2)+y_(2)^(2)-2sqrt2 y_(2) = r^(2)-2" "...(iii)`
`x_(3)^(2)+y_(3)^(2)-2sqrt2 y_(3) = r^(2) - 2 " "...(iv)`
oN substracting Eq. (ii) from Eq. (iii), we get
`P_(1)-sqrt2q_(1)=0`
where `p_(1)=x_(2)^(2)+ y_(2)^(2)-x_(1)^(2)-y_(1)^(2)`,
`q_(1)=y_(2)-y_(1)`
On substracting Eq. (ii) from Eq. (iv), we get
`p_(2)-sqrt2q_(2)=0`
where `p_(2)=x_(3)^(2)+y_(3)^(2)-x_(1)^(2)-y_(1)^(2), q_(2)=y_(3) - y_(1)`
Now, `p_(1) , P_(2), q_(1),q_(2)` are rational numbers. Also, either `q_(1) ne 0 or q_(2)ne 0`. If `q_(1) ne 0`, then `sqrt2=p_(1)//q_(1) and " if" q_(2) ne 0` then `sqrt2 = p_(2)//q_(2)`. In any case `sqrt2` is a rational number. This is a contradiction.