两个不同的圆的最多相交两次

Suppose that we have two circles in the same plane. We will prove that the circles intersect at most twice.

Let the radii of the circles be $R_1$ and $R_2$. Let the distance between the circle centers be $d$.

Notice that $d$, $R_1$, and $R_2$ are all fixed.

Let $O_1$ and $O_2$ be the centers of the first and second circle, respectively. Let $P$ be a point lying on both circles.

Since $P$ lies on both circles, we have: $$O_1P=R_1$$ $$O_2P=R_2$$ If $d>R_1+R_2$, then obviously there is no solution.

Next, if $d=R_1+R_2$, then the point $P$ is the point of tangency of the two circles. Here, there is only one solution.

Finally, we have the nontrivial case: $d<R_1+R_2$. In this case, $O_1$, $O_2$, and $P$ form a nondegenerate triangle.

But wait! The lengths of $O_1P$, $O_2P$, and $O_1O_2$ are all fixed!

This means that, by SSS congruence, the triangles $\triangle O_1O_2P$ are congruent for all points $P$.

This also means that the angle $\angle O_1O_2P$ is congruent for all points $P$. This restricts $P$ to two rays symmetric across $O_1O_2$ and having endpoint at $O_2$. Taking into account the distance requirement $O_2P=R_2$ yields two points, one per ray. Thus, there are only two possible points for the circles to intersect.

If the circles are noncoplanar, then the fact that they intersect in at most two points is even more obvious.

First of all, the point must lie in both planes of the circles.

Since two planes meet in a line, all points of intersection must lie on a line.

A point and a line only meet in a maximum of two points, so the intersection of the circles is also at most two points.

Thus, we have proven that two circles meet in at most two points.