What is the geometry of a certain two dimensional set? Let $(x_1,y_1),(x_2,y_2),$ and $(x_3,y_3)$ be three distinct points in $\mathbb{R}^2$. What is the geometry of the following set?
$$\{(x,y)\in\mathbb{R}^2\mid det\begin{pmatrix} 1 & x_1 & y_1 & 
x_1^2+y_1^2 \\ 1 & x_2 & y_2 & x_2^2+y_2^2 \\ 1 & x_3 & y_3 & x_3^2+y_3^2 \\ 1 & x & y & x^2+y^2\end{pmatrix}=0\}$$ Immediately, I notice that these four vectors are collinear, meaning $\langle1,x,y,x^2+y^2 \rangle$ is in the cube formed by the other 3 vectors (for now I'm considering the case where the first 3 are linearly independent). Additionally, all of these vectors lie on the cube $$\langle 1,0,0,0\rangle+t\langle0,1,0,0\rangle+r\langle0,0,1,0\rangle+s\langle0,0,0,1\rangle$$I'm pretty sure the intersection of two cubes in $\mathbb{R}^4$ is a plane, which is what I'm looking for. I'm not very good at picturing objects in higher dimensions, but my intuition tells me this plane will be a triangle between the three points. (If one of the first 3 vectors were a linear combination of the others, then this would form a line.) However, from here I'm lost as to how I'm supposed to find the equation of this plane and interpret it.
 A: you can observe that 
$det\begin{pmatrix} 1 & x_1 & y_1 & 
x_1^2+y_1^2 \\ 1 & x_2 & y_2 & x_2^2+y_2^2 \\ 1 & x_3 & y_3 & x_3^2+y_3^2 \\ 1 & x & y & x^2+y^2\end{pmatrix} =$
$det\begin{pmatrix}  x_1 & y_1 & 
x_1^2+y_1^2 \\  x_2 & y_2 & x_2^2+y_2^2 \\  x_3 & y_3 & x_3^2+y_3^2 \\ \end{pmatrix}-x det\begin{pmatrix} 1 &  y_1 & 
x_1^2+y_1^2 \\ 1  & y_2 & x_2^2+y_2^2 \\ 1  & y_3 & x_3^2+y_3^2 \\ \end{pmatrix}+ $
$+y det\begin{pmatrix} 1 &  x_1 & 
x_1^2+y_1^2 \\ 1  & x_2 & x_2^2+y_2^2 \\ 1  & x_3 & x_3^2+y_3^2 \\ \end{pmatrix}-$
$-(x^2+y^2)det\begin{pmatrix} 1 &  x_1 & 
y_1 \\ 1  & x_2 & y_2 \\ 1  & x_3 & y_3 \\ \end{pmatrix}$
So the set of  solutions it is a conic of the plane in the case in which the determinant 
$det\begin{pmatrix} 1 &  x_1 & 
y_1 \\ 1  & x_2 & y_2 \\ 1  & x_3 & y_3 \\ \end{pmatrix}\neq 0$
that means that the three points are in a general position, namely the three vector have all different directions.
You can observe that:
1: The conic is a cirumference;
2: all the three points are in the conic because the matrix has two raws equal and so the determinant of the matrix is zero;
But the three points are in general position so the conic is the only circumference of the plane that contains the three points. 
(Infact it is simple verify that there exists only one cirumference that contains three fixed distinct points in general position)
