Rational points on a circle with a center having non rational coordinates Reflection of a rational point with respect to line having its equation with rational coefficient
gives a rational point
What is the number of rational points on a circle with centre $(\sqrt 2, \sqrt 3)$ and with any radius
In this the centre is not rational then also how they decided the answer as 'at most 1'
Original source : https://i.stack.imgur.com/3aMvD.gif
 A: Suppose for the sake of contradiction that the circle contains two rational points.
If we have two rational points $A(x_1, y_1)$ and $B(x_2, y_2)$ on the circle, then the perpendicular bisector of $AB$ has equation
$$y - \frac{y_1 + y_2}{2} = -\frac{x_2 - x_1}{y_2 - y_1}\left(x - \frac{x_1 + x_2}{2}\right)$$
which can be written as $ax + by = c$ for rational $a, b, c$.
But this perpendicular bisector must pass through $(\sqrt2, \sqrt3)$, the center of the circle, so we have $a \sqrt 2 + b \sqrt3 = c$. This is a contradiction: there is no rational relation between $\sqrt 2$ and $\sqrt 3$.
Therefore at most one rational point can lie on the circle.
A: Let $(a,b)$ and $(c,d)$ be distinct rational points on the circle.
$$\begin{array}{rcl}
(a-\sqrt2)^2 + (b-\sqrt3)^2 &=& (c-\sqrt2)^2 + (d-\sqrt3)^2 \\
a^2 + b^2 - 2a\sqrt2 - 2b\sqrt3 &=& c^2 + d^2 - 2c\sqrt2 - 2d\sqrt3\\
a^2 + b^2 - c^2 - d^2 &=& 2(a-c)\sqrt2 + 2(b-d)\sqrt3\\
(a^2 + b^2 - c^2 - d^2)^2 &=& 8(a-c)^2 + 12(b-d)^2 + 8(a-c)(b-d)\sqrt6 \\
\sqrt6 &=& \dfrac{(a^2 + b^2 - c^2 - d^2)^2 - 8(a-c)^2 - 12(b-d)^2}{8(a-c)(b-d)}\\
\end{array}$$
Contradicting the fact that $\sqrt6$ is irrational, unless $a=c$ or $b=d$.

I believe you can deal with the cases:


*

*Case 1: $a=c$ and $b \ne d$

*Case 2: $a \ne c$ and $b=d$


To complete the proof.
