Let's assume we have two circles given by their origin point and radius: $$ (x_0, y_0, r_0) , (x_1, y_1, r_1) $$
Is it true, that for any such two circles, if there exists an intersection of the two, the largest distance between any two points located inside the intersection area is always no greater than the distance between the intersection points?
By a simple mspaint drawing, I managed to find a contradiction to this statement:
Two intersecting circles where the statement is false
But now I wonder, if we impose an additional restriction, such that
$$ r_0 = r_1 $$
does the statement become true for all such circles?
As long as each circle, no more than half of the circumference is in the intersection, you'll find the entire intersection fits inside the smallest circle containing the two intersection points. Because any points within this circle have a distance smaller than the distance between the intersection points, this is true in particular for points in the intersection. With equal radii, it's not possible for more than half of the circumference to be in the intersection.
Consider one half of the intersection area. Consider what it would look like for all possible radii of the circle on the opposite side. Starting with a circle of infinite radius, it would be a line, and as you make the radius smaller, it bulges out more, so the intersection area on that side gets strictly bigger when you shrink the radius, up until the point where the radius is too small for any circle to exist which goes through both points. The smallest possible circle is the one where the two points are opposite each other. So this is the one that maximises the intersecting area on that side. Hence, in particular, the intersecting area from whatever circle you started with is contained in it.
It's intuitively obvious that smaller circles with more curvature 'bulge out' more in a way that contains the 'bulge' of larger circles, but I don't know a neat proof of this.