<s>[INST] Linear Algebra: Looking at points M on a circle. Consider the points
$
M=\left\{ 
\begin{pmatrix}
0\\
2
\end{pmatrix}
,
\begin{pmatrix}
2\\
2
\end{pmatrix}
,
\begin{pmatrix}
1\\
0
\end{pmatrix}
,
\begin{pmatrix}
1\\
2
\end{pmatrix}
\right\}
$
in $\mathbb{R}^2$.
The points on a circle with radius r and center in $(a,b)^T\in\mathbb{R}^2$ is described as the solutions to the equation $(X-a)^2+(Y-b)^2=r^2$. This equation is equivalent with $2aX+2bY+(r^2-a^2-b^2)=X^2+Y^2$
Set
$$
\begin{pmatrix}
2a\\
2b\\
r^2-a^2-b^2
\end{pmatrix}
\in \mathbb{R}^3
$$
I have to show that the points in $M$ is on the circle if and only if $v$ is the solution to the linear equation system
$$
\begin{pmatrix}
0 & 2 & 1\\
2 & 2 & 1\\
1 & 0 & 1\\
1 & 2 & 1
\end{pmatrix}
\cdot x
=
\begin{pmatrix}
4\\
8\\
1\\
5
\end{pmatrix}
$$
I have already shown the way where we assume $M$ is on the circle and show that $v$ is then the solution to the linear equation system. However, i'm having trouble the other way around. How do i show that $M$ is on the circle if $v$ is the solution to the linear equation system? I have tried inserting the vector 
$$
\begin{pmatrix}
2a\\
2b\\
r^2-a^2-b^2
\end{pmatrix}
$$
in the equation instead of x, however, I don't see how this is supposed to help me.. 
 [/INST] Take a generic circle $\mathcal C_{a,b,r}=\{(X,Y)\in \mathbb R^2 \ | \ 2aX+2bY+(r^2-a^2-b^2) = X^2+Y^2\}$. 
We have:
\begin{equation}
M\subseteq \mathcal C_{a,b,r} \Longleftrightarrow 
\begin{cases}
2a (0) +2b (2) + (r^2-a^2-b^2) = 0^2+2^2 = 4\\
2a (2) +2b (2) + (r^2-a^2-b^2) = 2^2+2^2 = 8\\
2a (1) +2b (0) + (r^2-a^2-b^2) = 1^2+0^2 = 1\\
2a (1) +2b (2) + (r^2-a^2-b^2) = 1^2+2^2 = 5\\
\end{cases}
\end{equation}
Where the system means that the points in $M$ satisfies the equation of $\mathcal C_{a,b,r}$. Setting now 
\begin{equation}
v=\left(\begin{matrix}
2a\\
2b\\
r^2-a^2-b^2
\end{matrix}\right)
\end{equation}
we can rewrite the linear system in a more suitable way:
\begin{equation}
M\subseteq \mathcal C_{a,b,r} \Longleftrightarrow 
\begin{pmatrix}
0 & 2 & 1\\
2 & 2 & 1\\
1 & 0 & 1\\
1 & 2 & 1
\end{pmatrix}
\cdot v
=
\begin{pmatrix}
4\\
8\\
1\\
5
\end{pmatrix}
\end{equation}
That is your statement.

Moreover, if you try to resolve the system written above, you will discover that it does not have solution; hence it does not exist a circle contained every points of $M$. 
We can see that this result is coherent: infact the points $\left\{ 
\begin{pmatrix}
0\\
2
\end{pmatrix}
,
\begin{pmatrix}
2\\
2
\end{pmatrix}
,
\begin{pmatrix}
1\\
2
\end{pmatrix}
\right\}
\subseteq M$ are aligned, so we had to expect this result.
</s>