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Any three non collinear points lie on the circumference of a unique circle.


Any three non collinear points may be represented on a cartesian plane. For ease of calculation the For ease of calculation these may be translated and rotated without distorting dimensions so that points are represented by $\alpha (0,0), \beta (0,a)$ and $\gamma (b,c)$.

Three lines$(\alpha\beta, \alpha \gamma, \beta\gamma)$ may be drawn between these points.

The gradients of these lines are

$\alpha\beta = \frac{a}{0}$ (undefined)

$\alpha\gamma = \frac{c}{b}$

$\beta\gamma = \frac{c-a}{b}$

The mid points of these lines are

$\alpha\beta = (0,\frac{a}{2})$

$\alpha\gamma = (\frac{b}{2},\frac{c}{2})$

$\alpha\gamma = (\frac{b}{2},\frac{a+c}{2})$

Therefore, the gradients of the perpendicular bisectors of these lines (denoted by $'$) are

$\alpha\beta' = 0$ (undefined)

$\alpha\gamma' = \frac{-b}{c}$

$\beta\gamma' = \frac{b}{a-c}$

The equations of these perpendicular bisectors are

$\alpha\beta': y = \frac{a}{2}$

$\alpha\gamma': y = \frac{-bx}{c}+k$


$\beta\gamma': = y=\frac{bx}{a-c}+k$


Intersections of the Bisectors














These two perpendicular bisectors intersect at the same point. Points on a perpendicular bisector are equidistant from the two points. Therefore All three points are$(\alpha,\beta,\gamma)$ equidistant from the point of intersection. Therefore a circle may be drawn centred on the intersection which will pass through all three points.



It is possible to extend ths proof to collinear points by the following.

As the radius of a circle increases the curvature decreases.

As the $radius \to \infty, curvature \to 0$.

Something with 0 curvature is a straight line.

Therefore, collinear points lie on the circumference of an infinitely large circle.