Concyclicity in the Euclidean Plane

I want to address the question of how to determine whether or not four points are concyclic. In a previous post, this question reared its head and was solved by an introduction of the circumcentre of three of the points. In this post, I present a direct approach.

We prove the following:

Let $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$ be four points in $\mathbb{R}^2$ . Then these four points are concyclic if and only if

$\displaystyle \left| \begin{array}{cccc} x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \\ x_4^2+y_4^2 & x_4 & y_4 & 1 \end{array} \right|=0.$

For the proof, first suppose that the points $(x_i,y_i)$ all lie on the curve

$\displaystyle (x-x_0)^2+(y-y_0)^2=r^2.$

The four equations we obtain give a linear dependence of the columns in the $4\times 4$ matrix under consideration, hence its determinant is zero.

Conversely, suppose that the determinant in question is zero. Then there is a nonzero vector in the nullspace of our $4\times 4$ matrix. If its first entry is nonzero, we can normalise it to be of the form $(1,-2x_0,-2y_0,x_0^2+y_0^2-r^2)^T$ and we obtain the circle that the four points lie on. If the first entry of this vector is equal to zero, then the four points must lie on a line, which after is all is just a degenerate circle, QED.

This entry was posted on August 21, 2014 at 11:46 am and is filed under maths. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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