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
$\begin{pmatrix} \alpha & \beta & 1 \end{pmatrix}\cdot v=\alpha^2+\beta^2$.
Solution: I assume that $(\alpha, \beta)^T \in \mathbb{R}^2$ is a point on the circle $(X-a)^2 + (Y-b)^2 = r^2$ and want to show that this implies $ \begin{pmatrix} \alpha & \beta & 1 \end{pmatrix} \cdot v = \alpha^2 + \beta^2 $. As $(\alpha, \beta)^T$ is a point on the circle is a solution to the equation $2aX+2bY + (r^2-a^2-b^2) = X^2 + Y^2$. We calculate and have that $$ \begin{pmatrix} \alpha & \beta & 1 \end{pmatrix} \cdot v = \begin{pmatrix} \alpha & \beta & 1 \end{pmatrix} \cdot \begin{pmatrix} 2a \\ 2b \\ r^2-a^2-b^2 \end{pmatrix} = 2a\alpha + 2b\beta + \left(r^2-a^2-b^2 \right) = \alpha^2 + \beta^2 $$ Can you verify if this way is correct? I think it is. The other way I am a little bit unsure about. I read another post with the same question and they said to use that $\alpha^2 + \beta^2 = \alpha \cdot \alpha + \beta \cdot \beta = |\alpha|^2 + |\beta|^2 = |\alpha + \beta|^2 = \langle \alpha + \beta, \alpha + \beta \rangle$ by using the Pythagores and and the fact that the dot product of is the square of the norm. But how do I go from here?
My own solution, without using the above, were to assume that $\begin{pmatrix} \alpha & \beta & 1 \end{pmatrix}\cdot v=\alpha^2+\beta^2$ is met and want to show that this implies that $(\alpha, \beta)^T \in \mathbb{R}^2$ is a point on the circle $(X-a)^2 + (Y-b)^2 = r^2$. But as the first is met can I just use this to say that $(\alpha, \beta)^T$ must be a point on the circle $(X-a)^2 + (Y-b)^2 = r^2$. But by doing this have I just assumed what I want to show?
Thanks in advance.