Calculating position of overlapping ovals to create a certain sized center gap

Question

I'm trying to solve a problem how to position five circles in longitude and latitude so that four of them overlap so that the fifth one in the middle is exposed by a certain margin.

I drew the diagram below quickly to illustrate what I'm trying to do. Each circle is 10 meters diagonally (or miles, whichever you prefer). The center one is fixed to its position. I'm trying to figure out how to calculate the position of four others so that there is approximately 1m2 of the center circle exposed. Essentially the four circles should create a 1m2 rectangle in the middle (obviously it's not an actual rectangle due to the curvature).

The positioning itself isn't that hard but what's throwing me off is how to deal with latitude and longitude where length of a degree is different depending on the coordinates.

All help is appreciated a lot because it's been a long time since I had to work with this type of math, and I can honestly say I'm completely lost :/

Answer 1

Let's use $(0,0)$ the center of the green circle. To fit a $1m\times 1m$ box in the middle, the distance from that center to any of the circles is $0.5m$. The radius of each of the purple circles is $5m$. So the distance from the origin to each circle is then $5.5m$. Therefore the positions are $(0,5.5)$, $(0,-5.5)$, $(5.5,0)$, and $(-5.5,0)$. The visible area is slightly larger than $1m^2$. If you want to have exactly that area, you need to do an integral.

Answer 2

Observe that the center of the four circles of radius $r$ form a square. So, it is convenience to place the four circles in terms of the distance $d$ between the adjacent centers.

If the overlapping area of any two circles is $p$, then the exposed area in the middle is

$$A= d^2 - \pi r^2 + 2p$$

It is known that the overlapping area $p$ between two circle of radius $r$ and at a distance $d$ between their centers is given by

$$p = 2r^2\cos^{-1}\left( \frac{d}{2r} \right) - \frac{1}{2}d\sqrt{4r^2-d^2}$$

So, in order to achieve 1$m^2$ for the exposed area, just set up the following equation,

$$1 = d^2 - \pi r^2 + 4r^2\cos^{-1}\left( \frac{d}{2r} \right) - d\sqrt{4r^2-d^2} $$

Given that $r= 5\text{m}$, the distance $d$ among the four circle centers can be solved, albeit maybe numerically.

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In terms of latitudes and longitudes, assuming the earth radius is $R$ and arbitrary center circle coordinates $(\theta, \phi)$, the right circle is to be placed at

$$\text{latitude}:\>\> \theta; \>\>\>\> \text{longitude}:\>\> \phi + 90^\circ\times \frac{\sqrt{2}d}{\pi R\cos\theta} $$

Similar positions can be derived for other three circles.