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Prove that three points are enough to draw/define one and only one circle, how would this be done?

J. M. ain't a mathematician
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LinAlgStudent
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3 Answers3

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The set of points that are equidistant from two points $A,B$ is a straight line, right? Precisely, the line that goes through the midpoint of the segment $AB$ and is perpendicular to it. Let's call this line $l(A,B)$.

Given 3 non-collinear points $A,B,C$, the lines $l(A,B)$ and $l(A,C)$ are not parallel, because the lines $AB$, $AC$ are not parallel, and therefore meet in exactly one point. This point is equidistant from $A,B,C$, and is therefore the only such point. It is the center of the unique circle that goes through these three points.

Andrés E. Caicedo
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    Note that three points which are collinear still define a circle in an appropriately generalized sense (the line through them) of infinite radius. – Qiaochu Yuan Jan 07 '11 at 01:50
  • And if we allow generalised circles then we may as well use the machinery of Möbius transformations: Take the three points to $0, 1, \infty$ (by 3-transitivity), and then it's clear there is exactly one generalised circle through those three points. But Andres's argument is easier to understand geometrically. – Zhen Lin Jan 07 '11 at 03:25
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    "the lines $l(A,B)$ and $l(A,C)$ are not parallel, because the lines $AB$, $AC$ are not parallel" ... assuming Euclid's Fifth, of course. :) – Blue Jan 07 '11 at 07:08
  • @Blue How does fifth postulate prove that? – user103816 Jul 20 '15 at 09:56
  • @user103816: It's easier to see that the lack of Euclid's Parallel Postulate causes the statement to be false. In hyperbolic geometry, for instance, let's take $P$ somewhere in the interior of $\angle BAC$, and let $M$ and $N$ be the midpoints of $AB$ and $AC$. Now, if $\angle PAB$ and $\angle PAC$ are equal-to-or-larger-than the "angles of parallelism" for $AM$ and $AN$, then $\overleftrightarrow{AP}$ and $l(A,B)$ are asymptotically-or-divergently parallel, as are $\overleftrightarrow{AP}$ and $l(A,C)$. Thus, $l(A,B)$ and $l(A,C)$ can indeed be parallel, even though $AB$ and $AC$ aren't. – Blue Jul 20 '15 at 11:12
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An expansion on the other answer.

More analytically (or explicitly), let $a$,$b$,$c$ be three points. We want to show there is only one point equidistant to all three.

Let $a = (a_1,a_2)$, $b = (b_1,b_2)$,$c = (c_1,c_2)$. Points $x = (x_1,x_2)$ equidistant to all three must satisfy,

$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2=(x_1-c_1)^2+(x_2-c_2)^2$$

Which is the system of equations

$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2$$ $$(x_1-b_1)^2+(x_2-b_2)^2=(x_1-c_1)^2+(x_2-c_2)^2$$

The solution to each equation is a line. Specifically, the solution to the first equation is the line

$$ (x_1-a_1)^2+(x_2-a_2)^2=(x_1-b_1)^2+(x_2-b_2)^2$$

$$ -2a_1x_1+a_1^2 -2a_2x_2+a_2^2=-2b_1x_1+b_1^2 -2b_2x_2+b_2^2$$

$$ x_1 = \frac{(-2b_2+2a_2)x_2+(b_2^2-a_2^2)}{-2a_1+2a_2}$$

Since two lines intersect at at most one place, the solution to the system is either a single point which defines the center of the circle, or there is no solution (the case of $a$,$b$,$c$ collinear). If we include the point at infinity as Qiaochu mentions then two lines always intersect at exactly one point (with parallel lines intersecting at infinity), and the circle will be through $a$,$b$,$c$ with infinite radius.

Otherwise the radius will be $\sqrt{(x_1-a_1)^2+(x_2-a_2)^2}$ and we have defined a circle.

AnonymousCoward
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Here is a simple proof:

Let A, B, and C be the three points. Draw a triangle between these three points. If you find the circumcenter of this triangle, you get the center of the circle containing all three points. Because there is only one circumcenter for each triangle, there is a defined circle for every three non-colinear points.

PotatoLatte
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