Let M be a collection of points $M=\begin{Bmatrix} \begin{pmatrix} 0\\2 \end{pmatrix},\begin{pmatrix} 2\\2 \end{pmatrix},\begin{pmatrix} 1\\0 \end{pmatrix},\begin{pmatrix} 1\\2 \end{pmatrix} \end{Bmatrix}$.
The points on a circle with radius r and center in $(a,b)^T\in\mathbb{R}^2$ is described by the equation $(X-a)^2+(Y-b)^2=r^2$
which is equivalent with the equation
$2aX+2bY+(r^2-a^2-b^2)=X^2+Y^2$
Now let $v=\begin{pmatrix} 2a\\2b \\r^2-a^2-b^2 \end{pmatrix}\in\mathbb{R}^3$
Show that $(\alpha,\beta)^T$ is a point on the circle described above if and only if
$\begin{pmatrix} \alpha & \beta & 1 \end{pmatrix}\cdot v=\alpha^2+\beta^2$.
$\begin{pmatrix} \alpha & \beta & 1 \end{pmatrix}\cdot\begin{pmatrix} 2a\\2b \\r^2-a^2-b^2 \end{pmatrix}=2a\alpha+2b\beta+r^2-a^2-b^2$
$\implies -((\alpha-a)^2+(\beta-b)^2)+r^2+\alpha^2+\beta^2$
$\implies -r^2+(\alpha-a)^2+(\beta-b)^2=0$
$\implies $ $\begin{pmatrix} \alpha & \beta & 1 \end{pmatrix}\cdot v=\begin{pmatrix} \alpha & \beta & 1 \end{pmatrix}\cdot\begin{pmatrix} 2a\\2a \\r^2-a^2-b^2 \end{pmatrix}=2a\alpha+2b\beta+r^2-a^2-b^2=\alpha^2+\beta^2$
So I believe I have managed to show it one way "=>", but I am stuck on the other way "<=" of the proof.