# Area of Overlapping Circles a Convex Function?

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

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.