The following problem has been bothering quite a while. I guess there is a gap in my school knowledge of geometry, but I do not know how to show that:
Prove or disprove that given two circles with centers $O_1, O_2$ of the same radius $r$, which intersect at two points $A, B$, the intersection of the circles is entirely contained in the circle with the center $M$ at the midpoint of the line $O_1O_2$ and radius $AM$.
In the picture below: $M$ would be the origin, and $A$ and $B$ are two points, where the circles intersect. The desired circle that is supposed to contain the intersection is drawn in green.
Best Answer
I think one possible way to approach this is to prove that the lines which constitute the area of the intersection of the two circles are within the derived circle. So the logic would be that, if the two curves are within the circle, the intersection is contained; otherwise it is not.
With the help of the coordinate system, and to simplify calculation, we can draw two circles as you did, where the centres are on the $x$-axis, both circles are symmetrical about $y$-axis, with individual radius being $r$, and distance between the two radii being $d$. We can obtain three functions for each circle: $$(x+a)^2+y^2=r^2$$ $$(x-a)^2+y^2=r^2$$ $$x^2+y^2=r^2-a^2$$
From the three equations, we compare the $x$ value of these three functions for $-\sqrt{r^2-a^2}≤y≤\sqrt{r^2-a^2}$, and it is then easy to see that the two curves are in between the two half circles, i.e. the two curves are contained in the circle. Thus the intersection is contained in the circle.