Area of Overlapping Circles a Convex Function?

Question

All,

I'm not sure how to phrase this. I've calculated the area of overlap of two circles of radius 1, with the first circle centered at $x=0$ and the second at $x=2d$ as $$A = 2\arccos(d) - 2d\sqrt{1-d^2}$$

I'm pretty confident that this is correct. The circles touch at $x=d$, the midpoint of the centers. This is a convex function for $d\in[0,1]$, which I'm struggling to understand. That means that if I slide a circle from $x=2d$ to $x=-2d$ there will be a cusp in the graph of area, which seems unintuitive?

Is this true? I feel like it should be a smooth curve...

Here is a graph I've generated of this case in MATLAB GRAPH

Answer 1

Let's analyse the derivative of our area function at a given $d>0$. As you noted, the circles intersect at $x=d$. Let $h$ denote the height difference of the two intersection points. If we move the right disk a tiny $\epsilon>0$ to the right, then the change in area is approximately $\epsilon h$. This is because at any fixed horizontal line through the intersection, the change in width of the intersection is exactly $\epsilon$ (or approximatly, but only when the horizontal line is so close to the edge that the new intersection does not go through the line.) Since $d>0$, the area decreases, so the derivative is exactly $-h$. However if $d<0$, then the area would have increased, so the derivative would have been exactly $+h$. I believe this is enough to explain the convexity and the cusp in the graph.