Euclid's Elements were a great work, but in modern standards it is not totally rigorous. One of its biggest flaws is right on Proposition 1, where he doesn't prove that the two circles intersect, in order to build an equilateral triangle.

The problem of intersecting circles also comes up in the proofs of simple facts, like the existence of midpoints for every line segment.

To prove this we need additional axioms. Modern approaches have fixed this issue. Even though it is seemingly trivial, I don't seem to be able to prove it. Even Hilbert's Foundations of Geometry skipped this result.

So, my question is:

Using a modern axiomatic approach, like Hilbert's axioms, how could one prove that, given the points $A$ and $B$, the circles centered respectively in $A $ and $B $ and radius $\overline{AB}$ intersect in (at least) two points?

I hope you can enlighten me. Maybe this is so simple I can't see it, or I haven't looked up in the right place.

  • $\begingroup$ Do you mean to say that the circles intersect in at least one point or at most two points? $\endgroup$ – pseudoeuclidean Dec 29 '16 at 16:09
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    $\begingroup$ I mean at least two points, because I can find proofs for at most two points (and this holds in general, while what I want does not). $\endgroup$ – J. C. Dec 29 '16 at 16:49

This is discussed in tremendous detail in Greenberg's Euclidean and Non-Euclidean Geometry, which includes the "Circle-Circle Continuity Principle":

If a circle $\gamma$ has one point inside and one point outside another circle $\gamma'$, then the two circles intersect in two points.

Greenberg also discusses the related "Line-Circle Continuity Principle" and "Segment-Circle Continuity Principle". His treatment shows that the first of these principles implies the other two. He also writes (p. 131, 4th edition):

You may wonder why we have called these three statements "principles" instead of "theorems" or "axioms". The latter two would be theorems if we assumed the first one (as we will later show), but we do not wish to call the first one an axiom because we wish to illuminate exactly where it is needed, and then we will ad it as a hypothesis.

Greenberg goes on to note (p. 137) that the circle-circle continuity principle can be proved from Dedekind's axiom:

Suppose that the set $\{l\}$ of all points on a line $l$ is the disjoint union $\Sigma_1 \cup \Sigma_2$ of two nonempty subsets such that no point of either subset is between two points of the other. Then there exists a unique point $O$ on $l$ such that one of the subsets is equal to a ray of $l$l with certex $O$ and the other subset is equal to the complement.

Greenberg notes that the proof of the Circle-Circle principle from the Dedekind axiom is proved by Heath in his commentary on the Elements.

I should add that I think (not quite positive) that Dedekind's axiom is a consequence of (perhaps is equivalent to?) Hilbert's "Axiom of Completeness". So it is not correct to say that Hilbert does not address this.

  • $\begingroup$ Thank you, this was exactly what I needed! Thanks for the references, I didn't know the name of this result, so I couldn't look it up properly. Now I know how to research further. As for the final part, this paper proves that Hilbert completeness and Dedekind completeness are indeed equivalent, so you are right in your assumption. $\endgroup$ – J. C. Dec 30 '16 at 1:34

Given two points $A$, $B$ in the plane there are the midpoint $M$ of the segment $[A,B]$ and the line $\ell$ orthogonal to $A\vee B$ through $M$. The congruence axioms guarantee that all points $P\in\ell$ have equal distance from $A$ and $B$. The point $M\in\ell$ has distance ${1\over2}|AB|$ from these two points, and points far out on $\ell$ have a distance larger than $|AB|$ from these two points. "Continuity" then implies the existence of two points $P$, $Q\in\ell$ with $|PA|=|QA|=|AB|$, hence also $|PB|=|QB|=|AB|$. It follows that the two circles mentioned in the question intersect at least in the two points $P$ and $Q$.

In this proof we have made use of some sort of "completeness" of the line $\ell$, but not of intuitive arguments about interior and exterior of circles, etc.

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    $\begingroup$ Depending on what axiomatic system you're using, it may not be clear that $M$ or $\ell$ even exist at all. In traditional Euclidean geometry, you construct $M$ and $\ell$ by intersecting the two circles... $\endgroup$ – Eric Wofsey Dec 29 '16 at 20:33
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    $\begingroup$ @Christian Blatter Thank you for your input, but at least the proof I know for the existence of midpoint uses this fact, so it would be circular. $\endgroup$ – J. C. Dec 30 '16 at 1:36

I happen to have a copy of "Geometry with Trigonometry" by Patrick D. Barry [published in 2001], so I just reviewed the start of the chapter on circles to see how this is handled. It turns out that this author approaches the subject by explicitly introducing real-numbered measurements of distances between points (and angle sizes etc), then uses the Triangle Inequality and Pythagorean Theorem to reason about certain points on the line. The gist is that he first locates the foot of the perpendicular from the circle center to the line; then he reasons (a) if the foot is on the circle, all other points are farther from the center than it, so the line intersects the circle in one point; (b) if the foot is outside the circle, so are all other points on the line, so the line doesn't intersect the circle; (c) if the foot is inside the circle, there are two points on the line whose distance from the foot are given by a square-root expression - here he uses the distance from the center to the foot and the radius and works the Pythagorean Theorem backward - then shows that the distance from the center to each point is equal to the radius, so the points must be on the circle (the line passing through the center is handled as a special case).

ADDENDUM: I just remembered that the question was about two circles, whereas the above argument deals with a line and a circle. The author doesn't explicitly show that, so far as I can see, but it probably follows from the above argument in short order, given the right construction. In any case, you can see how such arguments can be made.


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