For a circle, suppose we distribute $n$ dots evenly near the circumference. One dot has a center at $(1-r,0)$ and the adjacent dots have centers at $$((1-r)\cos(2\pi/n),\ \pm (1-r)\sin(2\pi/n))$$ The middle one goes as far to the left as $1-2r$, and the neighbors go as far to the right as $(1-r)\cos(2\pi/n)+r$ To avoid collinearity, we need \begin{align} (1-r)\cos(2\pi/n)+r &< 1-2r\\ n &< 2\pi/\arccos\left(\frac{1-3r}{1-r}\right) \end{align} Asymptotically, this means we can choose $n$ as large as $$\frac{\pi}{\sqrt{r}}\left(1-\frac{2r}{3}+O(r^2)\right)$$ which is a reasonable lower bound for the problem.

For a unit square, $[-1/2,1/2]^2$, the same set up also seems to work well. Instead of using one circle, we can try connecting four circular arcs which stay closer to the four sides. The rightmost arc will have a center at $(-h,0)$, a radius of $k=1/2+h-r$, and a dot with a center at $(1/2-r,0)$. The total number of dots on that arc will be twice the floor of $\alpha/\beta$ where \begin{align} -h+k\cos(\alpha)&=k\sin(\alpha)\\ -h+k\cos(\beta)&=1/2-3r \end{align} Empirically, we maximize $\alpha/\beta$ at $h=0$, which is to say that we might as well take the four circular arcs to be part of the same circle.

For a circle, suppose we distribute $n$ dots evenly near the circumference.

One dot has a center at $(1-r,0)$ and the adjacent dots have centers at $$((1-r)\cos(2\pi/n),\ \pm (1-r)\sin(2\pi/n))$$ The middle one goes as far to the left as $1-2r$, and the neighbors go as far to the right as $(1-r)\cos(2\pi/n)+r$ To avoid collinearity, we need \begin{align} (1-r)\cos(2\pi/n)+r &< 1-2r\\ n &< 2\pi/\arccos\left(\frac{1-3r}{1-r}\right) \end{align} Asymptotically, this means we can choose $n$ as large as $$\frac{\pi}{\sqrt{r}}\left(1-\frac{2r}{3}+O(r^2)\right)$$ which is a reasonable lower bound for the problem.

For a unit square, $[-1/2,1/2]^2$, putting all the dots on one circle still seems to work well.

We can try an alternative, putting dots on four circular arcs (black, blue, purple, red) each of which stays closer to one of the four sides.

The black arc will have a center at $(-h,0)$, a radius of $k=1/2+h-r$, and a dot with a center at $(1/2-r,0)$. 

The dashed lines show radii of the black arc; we call the smallest central angle $\alpha$ and the biggest central angle $\beta$. Then the total number of dots on that arc will be twice the floor of $\beta/\alpha$ where \begin{align} -h+k\cos(\beta)&=k\sin(\beta)\\ -h+k\cos(\alpha)&=1/2-3r \end{align} Empirically, we maximize $\beta/\alpha$ at $h=0$, which is to say that we might as well take the four circular arcs to be part of the same circle. In the diagram, that corresponds to spacing the dots evenly outside the light gray circle instead.

For a circle, suppose we distribute $n$ dots evenly near the circumference. One dot has a center at $(1-r,0)$ and the adjacent dots have centers at $$((1-r)\cos(2\pi/n),\ \pm (1-r)\sin(2\pi/n))$$ The middle one goes as far to the left as $1-2r$, and the neighbors go as far to the right as $(1-r)\cos(2\pi/n)+r$ To avoid collinearity, we need \begin{align} (1-r)\cos(2\pi/n)+r &< 1-2r\\ n &< 2\pi/\arccos\left(\frac{1-3r}{1-r}\right) \end{align} Asymptotically, this means we can choose $n$ as large as $$\frac{\pi}{\sqrt{r}}\left(1-\frac{2r}{3}+O(r^2)\right)$$ which is a reasonable lower bound for the problem.

For a unit square, $[-1/2,1/2]^2$, the same set up also seems to work well. Instead of using one circle, we can try connecting four circular arcs which stay closer to the four sides. The rightmost arc will have a center at $(-h,0)$, a radius of $k=1/2+h-r$, one dot on the arc with a center at $(1/2-r,0)$, and a total number of dots on the arc equal to $2\alpha/\beta$ where \begin{align} -h+k\cos(\alpha)&=k\sin(\alpha)\\ -h+k\cos(\beta)&=1-3r \end{align} Empirically, we maximize $2\alpha/\beta$ at $h=0$, which is to say that we might as well take the four circular arcs to be part of the same circle.

For a circle, suppose we distribute $n$ dots evenly near the circumference. One dot has a center at $(1-r,0)$ and the adjacent dots have centers at $$((1-r)\cos(2\pi/n),\ \pm (1-r)\sin(2\pi/n))$$ The middle one goes as far to the left as $1-2r$, and the neighbors go as far to the right as $(1-r)\cos(2\pi/n)+r$ To avoid collinearity, we need \begin{align} (1-r)\cos(2\pi/n)+r &< 1-2r\\ n &< 2\pi/\arccos\left(\frac{1-3r}{1-r}\right) \end{align} Asymptotically, this means we can choose $n$ as large as $$\frac{\pi}{\sqrt{r}}\left(1-\frac{2r}{3}+O(r^2)\right)$$ which is a reasonable lower bound for the problem.

For a unit square, $[-1/2,1/2]^2$, the same set up also seems to work well. Instead of using one circle, we can try connecting four circular arcs which stay closer to the four sides. The rightmost arc will have a center at $(-h,0)$, a radius of $k=1/2+h-r$, and a dot with a center at $(1/2-r,0)$. The total number of dots on that arc will be twice the floor of $\alpha/\beta$ where \begin{align} -h+k\cos(\alpha)&=k\sin(\alpha)\\ -h+k\cos(\beta)&=1/2-3r \end{align} Empirically, we maximize $\alpha/\beta$ at $h=0$, which is to say that we might as well take the four circular arcs to be part of the same circle.