Take a circle of radius $r$ with centre at the origin such that $r^2=N_1^2+N_2^2$ for $N_1,N_2\in\mathbb{N}$.
Consider a lattice coordinate $(a,b)$ such that $a\in(-r,-2)$ and define $b$ to be the smallest lattice $y$-ordinate such that
$$b>\sqrt{r^2-a^2} \qquad(\text{strictly}).$$
so that $P_1(a,b)$ is the closest lattice point (outside the circle) to $Q_1(a,\sqrt{r^2-a^2})$ in the vertical sense.
Define $P_2(a+1,c)$ in the same way so that
$$c>\sqrt{r^2-(a+1)^2},$$ so that $P_2(a+1,c)$ is the closest lattice point (outside the circle) to $Q_2(a+1,\sqrt{r^2-(a+1)^2})$ in the vertical sense.
Now using $\lambda\mathbf{x}+(1-\lambda)\mathbf{y}$ we have that points on the line segment $P_1P_2$ are given by:
$$(a+1-\lambda,\lambda b+c-\lambda c)\qquad\text{ for }\lambda\in[0,1].$$
I want to prove that none of these points are strictly inside the circle. This is equivalent to showing that:
$$b^2\lambda^2-2bc\lambda^2+c^2\lambda^2+2bc\lambda-2c^2\lambda+a^2-2al+c^2+\lambda^2+2a-2\lambda+1\geq r^2.$$
Any thoughts?
Obviously, this is not always so. Consider $r=\sqrt{1104},\;a=-33,\;b=4,\;c=9$, and look at the points $P_1(-33,4),\;P_2(-32,9)$, and the midpoint thereof.
Why obviously? Well, we have two points outside a circle, without any clear reason to stay far away from it; one should expect that with suitably chosen $r$ we might get them close enough to the circle so that the line segment will cut the circle.
As to how did I find that $r$: we want the two points to be just a tiny bit above the circle, i.e., $a^2+b^2$ and $(a+1)^2+c^2$ must be barely larger than $r^2$, ideally at $r^2+1$. Thus we want $r^2+1$ to be representable as a sum of two squares in at least two ways, so that these two ways use consecutive squares. From recent activity I recalled one sequence which contained a lot of such numbers. (Upd. Some of them satisfy your condition with squares, too.) The rest is trivial.