# Simple Geometric Proof of Intersecting Circles

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.

• Have you drawn a picture? – kccu Apr 23 at 13:21
• Yes, it looks obvious from the picture somehow. But I cannot put any coherent reasoning on paper. – Renat Sergazinov Apr 23 at 13:24
• Are they always on a horizontal line? That is, are the $y$ coordinates of their centers always the same? This problem is a LOT easier if so – Michael Stachowsky Apr 23 at 14:14
• The problem did not come with the coordinate system. But you can impose it anyway you like. – Renat Sergazinov Apr 23 at 14:32

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.