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

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up vote 18 down vote accepted

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.

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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 7 '11 at 1:50
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 7 '11 at 3:25
"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 7 '11 at 7:08
@Blue How does fifth postulate prove that? – user103816 Jul 20 '15 at 9:56
@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

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.

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