On smallest circles and ellipses containing at least n integer lattice points Continuing from On circles and ellipses drawn on an infinite planar square lattice, let us record two broad questions: In what follows, "contains" means "either contains within or passes thru".

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*Given an integer n, to find the smallest circle that contains at least n integer lattice points.

For n =3 and 4, the required circles have same radius =2. Is there any parallelogram lattice for which the smallest circle that contains at least n lattice points n has a unique radius for every value of n?


*Given an integer n, to find the smallest perimeter ellipse that contains at least n integer points (that there is no closed expression known for perimeter of ellipse need not make this question invalid).

Only for n <=3, the required ellipses seem degenerate. What can one say about sequences of successive values of n for which the least perimeter ellipses containing at least n integer lattice points are the same throughout? Is there any parallelogram lattice for which the least perimeter ellipse that contains at least n lattice points n is unique for every value of n?
Note 1:  Instead of at least n lattice points we could also look for containers that contain exactly n lattice points.
Note 2: Analogous questions are conceivable with other shapes of container  such as triangle, isosceles triangle etc. I am not sure if for all n, there are equilateral triangles that contain exactly n integer lattice points and if so whether all these triangles have unique sizes.
 A: If you want for each $n$ the minimal radius of a disk containing at least $n$ points we have the easily found pairs $(n,r^2)=(1,0),(2,\frac14),(4,\frac12),(5,1),(6,\frac54).$
At this point I would be tempted to say that the center can  be taken to be $(\frac{a}2,\frac{b}2)$ for $a,b$ integers. But it turns out I would be wrong. I'm not sure how much is known.
The circles shown below for $(6,\frac54),(9,2)$ and  $(12,\frac52)$ do have centers of this form. However a slight enlargement of the first can be shifted to give $(7,\frac{25}{16})$ shown here with center $(\frac34,3).$
And the circle for $n=12$ can be enlarged to give circles for $(n,r^2)=(13,\frac{169}{50})$ and $(14,\frac{65}{18})$ shown here with centers $(\frac{43}{10},\frac{43}{10})$ and $(\frac{25}6,\frac{17}{2}).$

That is enough data to find that  sequence $$  {1, 2, 4, 5,6},  7, {9}, {12}, 13, 14 \cdots$$ and some  illustrations.


LATER A slightly new question got added. Let $r_n$ be the minimal radius of any disk $D_n$ containing exactly $n$ (usual) lattice points. Then $r_3$ is undefined and $r_8=\frac32>\sqrt{2}=r_9.$ If $r_{10}$ and $r_{11}$ are both defined, then they are strictly between $r_9$ and $r_{12}.$ I wouldn't guess which is larger.


A disk of type $D_n$ will have a rational center: If it has two antipodal lattice points on the boundary the center has half integer coordinates. Otherwise there will be at least $3$ lattice points on the boundary (forcing a rational center.) Otherwise we could slightly shrink and shift to get the same points contained and a smaller radius.

For an appropriate fixed center $(x,y)$ (say $x,y,\frac{x}{y}$ all irrational) it should be the case that every disk with that center has at most one lattice point one the boundary. A disk
If we eliminate a certain countable set of lines from $\mathbb{R}^2$ "most" points will remain and for any one, no disk centered at it has more than one rational point on the boundary.
