Two Circles Can Have At Most One Common Chord? (IMO)

I am currently working to understand a combinatorics problem. Within the proof they state that "Two circles can have at most one common chord ..." and I do not understand why this is true. I've included a screenshot from my textbook.

The book then goes on to finish the proof. Now I can follow everything before the last paragraph, but have no idea why two circles can have at most one common chord (last full sentence of last paragraph).

Question: Can you please provide some explanation for why two circles can have at most one common chord?

• Can you think of an example of two circles intersecting with say 2 common chords? That attempt should give you an answer. Commented Oct 19, 2016 at 10:13
• Failure to dream up a counterexample is not normally considered a rigorous proof. Commented Oct 19, 2016 at 14:22
• No. But it can be an excellent start.
– John
Commented Oct 19, 2016 at 16:42
• If two circles touch at a point, wouldn't a multitude of lines pass through that point and touch both circles? Commented Oct 19, 2016 at 22:35
• Two circles * can* have more than one common chord but then they are the same circle (and have infinite chords in common) Commented Oct 20, 2016 at 4:01

If two circles have two chords in common, then they must have at least three points in common.

A circle can be completely determined by three points, as there are three unknowns: the $x$ and $y$ coordinates of the circle's center, and the radius.

If you want something more concrete: if you have two points $(x_0,y_0)$ and $(x_1,y_1)$, then you know that the center of the circle must be equidistant from both; so, there is an infinite line of points to choose from for the center, and choosing the center completely determines the radius.

When you introduce a third point $(x_2,y_2)$, you can now note that the center of the circle must be equidistant from $(x_0,y_0)$ and $(x_1,y_1)$, and must also be equidistant from $(x_0,y_0)$ and $(x_2,y_2)$. This gives two lines, whose intersection (if such exists) must be the center. Assuming there is a good point for the center, the radius must be the common distance from this point to all three $(x_i,y_i)$ pairs.

• I don't think the last point is complete. If the x & y of the circle's center are distinct unknowns, then three points actually have six knowns (the x & y of each). Thus one might expect that two points (four knowns) could uniquely identify a circle. As this is not the case, it doesn't immediately follow that bumping this to three points (six knowns) is sufficient to determine it. -- Contrast with a line segment, which has four unknowns (the x & y coordinates of the endpoints) but is determined either by two points (the end points) or can't be defined by an infinite number of internal points.
– R.M.
Commented Oct 19, 2016 at 15:05
• A data point is, in this case, a relationship between an $x$ coordinate and a $y$ coordinate, not the coordinates themselves. The argument isn't perfect, but it is illustrative. Commented Oct 19, 2016 at 15:09
• If you want something more concrete: if you have two points $(x_0,y_0)$ and $(x_1,y_1)$, then you know that the center of the circle must be equidistant from both; so, there is an infinite line of points to choose from for the center; radius is then determined. When you introduce a third point $(x_2,y_2)$, you can now note that the center of the circle must be equidistance from $(x_0,y_0)$ and $(x_1,y_1)$, and must also be equidistant from $(x_0,y_0)$ and $(x_2,y_2)$. This gives two lines, whose intersection (if such exists) must be the center, and you can immediately determine the radius. Commented Oct 19, 2016 at 15:10

Two distinct circles will intersect in at most two points. If they intersect in two points, they have a single common chord, namely the line segment joining those two points. If they intersect in fewer than two points, they have no common chords.

• Well, I think the OP was asking why they have at most two points in common... Commented Oct 19, 2016 at 3:48

Two circles can have at most two common points (they can intersect at 2 points at most). Hence they can have only one common chord.I hope this image is illuminating.

A circle is defined by three non-collinear points. Two chords are defined by at least three points, so by taking two chords of a circle, you take at least three points, which uniquely defines your circle.

Suppose the equations of two circles in a Cartesian Coordinate plane are given by:

$$A$$ ≡ {($$x$$,$$y$$) : A = $$x^2$$ + $$y^2$$ + $$2gx$$ + $$2fy$$ + $$c$$ = $$0$$}

$$B$$ ≡ {($$x$$,$$y$$) : B = $$x^2$$ + $$y^2$$ + $$2g'x$$ + $$2f'y$$ + $$c'$$ = $$0$$}

Consider a point ($$α$$,$$β$$)∈ $$A$$$$B$$. The duplet ($$x$$,$$y$$)=($$α$$,$$β$$) must satisfy both $$A$$ and $$B$$, and thus must also satisy

A-B = $$2(g-g')x$$ + $$2(f-f')y$$ + $$(c-c')$$ = $$0$$

The above represents the equation of a straight line. Thus, any point satisfying both $$A$$ and $$B$$ must also satisfy $$L$$ ≡ {($$x$$,$$y$$) : $$A-B$$ = $$2(g-g')x$$ + $$2(f-f')y$$ + $$(c-c')$$ = $$0$$ }, and thus must demarcate an intersection between $$L$$ and either of $$A$$ or $$B$$.

However, we know that any arbitrary line may intersect a circle at a maximum of two distinct points. Thus, the circles intersect at a maximum of two distinct points (there can be a maximum of two solutions to a system of a linear and a quadratic equation in two variables), which demarcate a single chord. Infact, $$L$$ represents the equation of the chord passing through the points of intersection of $$A$$ and $$B$$