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The homogenous representation of a circle is given by $x^2 + y^2 + 2gxz + 2fyz + cz^2 = 0$ (or, equivalently, if we set $z=1$, $x^2 + y^2 + 2gx + 2fy + c = 0$). Now, given 3 points (in a homogenous form), we can solve a system of linear equations and retrieve the unknowns $f$, $g$ and $c$.

This is all very nice (because of linear algebra), but what do these unknowns actually represent with respect to the circle? Which of these numbers represent the x and y coordinates of a circle and which one represents the radius?

Apparently, $-f$ and $-g$ would be the $x$ and $y$ coordinate of the center of the circle? Why is that the case? I would like to see a proof/derivation of it. Also, what is the radius then?

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  • $\begingroup$ @WillOrrick I believe this is the homogenous equation of a circle (I took it from Wikipedia). $\endgroup$
    – user168764
    May 28, 2020 at 13:57
  • $\begingroup$ @WillOrrick Yes, sorry. I guess we can ignore z if we set it to 1. Given we are working in homogenous coordinates, we can do that. $\endgroup$
    – user168764
    May 28, 2020 at 14:03
  • $\begingroup$ Yes, projective coordinates are not relevant to your question. Try completing the squares in both the $x$ and $y$ terms. $\endgroup$
    – Will Orrick
    May 28, 2020 at 14:05
  • $\begingroup$ @WillOrrick I don't understand your suggestion. What do you mean by "completing the squares in both the x and y terms"? If you can provide a formal answer to my question, I would appreciate. $\endgroup$
    – user168764
    May 28, 2020 at 14:05

1 Answer 1

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$$ \begin{aligned} 0&=x^2+y^2+2gx+2fy+c\\ &= x^2+2gx+g^2+y^2+2fy+f^2+(c-g^2-f^2)\\ &=(x+g)^2+(y+f)^2-(f^2+g^2-c) \end{aligned} $$ This equation says that the squared distance of the point $(x,y)$ from the point $(-g,-f)$ is $f^2+g^2-c$, which describes a circle centered at $(-g,-f)$ with radius $\sqrt{f^2+g^2-c}$.

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  • $\begingroup$ Ok, this is a straight acceptance (no, not yet lol), but just one clarification. The equation is homogenous, so how come that those represent the center and radius in Euclidean space? $\endgroup$
    – user168764
    May 28, 2020 at 14:14
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    $\begingroup$ The real projective plane contains the Euclidean plane. The only "extra" that it contains is the line at infinity (on which all the points at infinity lie). So the only place at which you might see new geometric features when moving to projective coordinates is at infinity. The finite part of the geometry remains the same. In the case of the circle, nothing new happens at infinity either, if you stick with real coordinates. The reason is that the line at infinity is described by $z=0$, and for your equation $z=0$ implies that $x=y=0$ as well. $\endgroup$
    – Will Orrick
    May 28, 2020 at 14:25

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