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On Wikipedia there is a Cartesian formula for the circumcenter of a triangle. That is, given points $A$, $B$ and $C$ in $\mathbb{R}^2$, find point $U$ such that $d(A,U)=d(B,U)=d(C,U)$. The formula, as stated on Wikipedia, is very algebraic, but I found that it can be rewritten as follows.

Define $X:=(A_x,B_x,C_x)$, $Y:=(A_y,B_y,C_y)$, $V:=(1,1,1)$ and $L:=(\|A\|^2,\|B\|^2,\|C\|^2)$. Then define the matrices $M_D:=(V,X,Y)$, $M_X:=(V,X,L)$ and $M_Y:=(V,L,Y)$. Then we have $$U=(|M_Y|,|M_X|) / (2|M_D|).$$

As I am writing a program, I very much enjoy this elegant form. However, such a nice formula needs a nice explanation if you ask me. Can anyone come up with one? I am hoping for an explanation that does not rely on any of the algebraic properties of the determinant and only uses its geometric interpretation based on measures.

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  • $\begingroup$ I’m not sure that there’s any particular geometric insight attached to your rearrangement of the formula, but I’d be quite interested in seeing one. You’ve basically taken a particular set of slices through the matrix in the determinant form of the equation of a circle through three fixed points. $\endgroup$ Commented Nov 14, 2018 at 20:01
  • $\begingroup$ Sorry, I had a mistake. I fixed it, though. I am confusing my own notation with the Wikipedia notation :/ $\endgroup$ Commented Nov 14, 2018 at 20:01
  • $\begingroup$ The Wikipedia article is awfully complicated : it is the perfect example of mathematical ill-writing ! The key concept behind the formula you give is barycentric coordinates that are very ill-explained there. I advise you to see them explained in a plain manner in the answer I gave (already 2 years ago) to this question : math.stackexchange.com/q/2106662 $\endgroup$ Commented Jan 26, 2019 at 23:26

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Actually it is really clever. Look what they do on Wikipedia. They are clever enough to point that the distance between the circle's center and each corner on the polygon should be equal ( = the radius). In the next step they form a system of equations. You will have to solve 2 systems of equations: one for x and one for y. In each system you will have only one unknown. There are countless way to solve numerically system of equations.

There are determinants used on Wikipedia. They compute the main determinant, and later "partial determinants". The end result is division of "partial determinant"/"main determinant". It is not good way for numerical methods. You had better stick to other ways of solving this system of equations. I am sorry, but I don't know the native english term for "partial determinant" and "main determinant".

If you have only one type of polygon to crank out, you may be better off with derivation of your formulas, like you are trying to do now. However, if you want to do software, which is going to work with all cyclic polygons, you should do this approach with solving of system of equations.

Good luck!

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    $\begingroup$ "It is not good way for numerical methods" : $3 \times 3$ determinants are not very harmful....and I think the case where the main determinant is close to 0 can be tested beforehand and corresponds to almost flat triangles... $\endgroup$ Commented Jan 26, 2019 at 23:20

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