Solution. First, we will prove that on any circle of the given family, there is no more than one point with integer (and even rational) coordinates. By contradiction, suppose that the rational points $M_{1}\left(x_{1}, y_{1}\right)$ and $M_{2}\left(x_{2}, y_{2}\right)$ lie on a circle of the given family. Then $\left(x_{1}-\sqrt{2}\right)^{2}+\left(y_{1}-\sqrt{3}\right)^{2}=\left(x_{2}-\sqrt{2}\right)^{2}+\left(y_{2}-\sqrt{3}\right)^{2}$. Therefore, $2\left(x_{1}-x_{2}\right) \sqrt{2}+2\left(y_{1}-y_{2}\right) \sqrt{3}=q$, where $q$ is a rational number. If $\left(x_{1}-x_{2}\right) \cdot\left(y_{1}-y_{2}\right) \neq 0$, then squaring the last equality, we get that $\sqrt{6}$ is a rational number, which is not true. If one of the differences, for example, $x_{1}-x_{2}$, is 0, then for $y_{1} \neq y_{2}$ we get a contradiction with the irrationality of $\sqrt{3}$. Therefore, there cannot be two different rational points on one circle. Next, note that inside a circle of small radius, there are no integer points (we can take a radius less than the distance from $M$ to the nearest integer point $A(1 ; 2))$. On the other hand, if we take a sufficiently large radius (for example, greater than 3000), then inside the circle there will be more than 2019 points. Since, by part a), when the radius is gradually increased, the number of integer points can only increase by one, there will inevitably be a moment when there are exactly 2019 points inside the circle.