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The idea is to avoid having to deal with numerical precision issues when evaluating whether Ptolemy's relationship holds, in a program. For example, in determining whether three points lie on a line, one can form a reduced matrix from the data points and simply determine its rank, thus avoiding any issues of numerical precision.

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4 Answers 4

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I’m not a geometer, but I think you could try the following:

Let ${(x_{i}, y_{i})}_{i=1}^{4}$ be your 4 points, then we can see that for some fixed center $(p,q)$ and radius $r$ each point must satisfy $(x-p)^2 + (y-q)^2 -r =0$ now as we have 4 points we can consider the 4 functions in $p,q,r$ defined by evaluating all 4 points in the rhs of the last equation. Now consider a sum of squares of those 4 functions and minimise it using some numerical optimization algorithm (or by hand), then if a minimizer $p,q,r$ is such that the objective function is zero then we found such a circle. If not, then it doesn’t exist.

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Four points $ABCD$ lie on a circle (in that order) if and only if $\angle ACB=\angle ADB$, that is if: $$ {(C-A)\cdot(C-B)\over|C-A|\ |C-B|}={(D-A)\cdot(D-B)\over|D-A|\ |D-B|}. $$

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For the four points to be on the same circle the point where perpendicular bisector of one pair of points meet the perpendicular bisector of the other pair should be equi-distance from all four points. We may check it algebraically if coordinates are given.

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Attempt:

Geometrically:

Let points $A,B,C$ and $D$ be given.

$A,B,C$, and $D$ lie on a circle $\iff$

The perpendical bisectors of $\overline{AB}$, $\overline {BC}$, and $\overline {CD}$ intersect in one point.

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