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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.

enter image description here

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?

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

5 Answers 5

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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.

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    $\begingroup$ 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. $\endgroup$
    – R.M.
    Commented Oct 19, 2016 at 15:05
  • $\begingroup$ 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. $\endgroup$ Commented Oct 19, 2016 at 15:09
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    $\begingroup$ 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. $\endgroup$ Commented Oct 19, 2016 at 15:10
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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.

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

Two circles with common chord

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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.

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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 $B$

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