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let A(A.x,A.y), B(B.x,B.y) and C(C.x,C.y) be some 3 points (ill use the .x and .y to represent the x,y coordinates of the points)

to start off, the line joining 2 points can be shown by equaling the gradient of 1 point to any point of the line, and the gradient between the 2 known points $$\frac{y-A.y}{x-A.x} = \frac{A.y-B.y}{A.x-B.x}$$

but for some reason, the same line can be represented using a determinant equaled to 0

$$ \begin{vmatrix} x & y & 1 \\ A.x & A.y & 1 \\ B.x & B.x & 1 \\ \end{vmatrix} =0 $$

this can be extended to a circle that passes through 3 points too as,

$$ \begin{vmatrix} x^2+y^2 & x & & y & 1 \\ A.x^2+A.y^2 & A.x & & A.y & 1 \\ B.x^2+B.y^2 & B.x & & B.y & 1 \\ C.x^2+C.y^2 & C.x & & A.y & 1 \\ \end{vmatrix} =0 $$

my question is, why does this work? i`ve not done linear algebra or such advanced topics yet, could someone explain why this works, and how to use this further?

Jean Marie
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Chinthana Nadun
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    For the circle, see the older Q&A here. The idea, as outlined in Piquito's answer, works for the line as well: the first row effectively encodes the equation of a given curve (as a linear function of its parameters), and the remaining rows encode the points required to uniquely specify a particular curve. – Semiclassical Aug 26 '24 at 06:54
  • This is related to Conformal Geometric Algebra through the relationship between geometric algebra, exterior algebra, and determinants. This idea works for arbitrary $n$-spheres. – Nicholas Todoroff Aug 28 '24 at 23:22
  • Also notice that your first determinant can be written $$\begin{vmatrix}x^2+y^2&x&y&1\1&0&0&0\A_x^2+A_y^2&A_x&A_y&1\B_x^2+B_y^2&B_x&B_y&1\end{vmatrix}.$$ (I've use subscripts instead of your dots.) This looks just like your circle-from-three-points determinant, and this is no accident: $(1,0,0,0)$ represents the point at infinity, and a line is just a circle passing through the point at infinity and two other points. – Nicholas Todoroff Aug 28 '24 at 23:31

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