<s>[INST] Equation of a circle in matrix form I have an equation $ \left( x-3 \right)^{2}+\left( y-3 \right)^{2}=9 $, and am trying to apply a matrix rotation of 180 degrees to it, however, I am having difficulty transferring the equation of the circle into matrix form so to complete the transformation.
Thanks
 [/INST] The equation $(x-3)^2+(y-3)^2=9$ is can be described using matrix and vector $[x \ y \ z]^{\text{T}}$ as follows. Then $z=1$.
$$
\begin{bmatrix}
x & y & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & -3 & \\
0 & 1 & -3 & \\
-3 & -3 & 9 &
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}
=0
$$
The rotation matrix is
$$
\begin{bmatrix}
\cos{\theta} & - \sin{\theta} & 0 & \\
\sin{\theta} & \cos{\theta} & 0 & \\
0 & 0 & 1 &
\end{bmatrix}
$$
$[x' \ y' \ z']^{\text{T}}$ is the transformed vector by the rotation matrix. When you want to translate 180 degree as $\theta = \pi$, the equation is decided below. 
$$
\begin{bmatrix}
x' & y' & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & -3 & \\
0 & 1 & -3 & \\
-3 & -3 & 9 &
\end{bmatrix}
\begin{bmatrix}
x' \\
y' \\
1
\end{bmatrix}
=0
$$
$$
\begin{bmatrix}
x & y & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 0 & 0 & \\
 0 & -1 & 0& \\
 0 & 0 & 1 &
\end{bmatrix}
\begin{bmatrix}
1 & 0 & -3 & \\
0 & 1 & -3 & \\
-3 & -3 & 9 &
\end{bmatrix}
\begin{bmatrix}
-1 & 0 & 0 & \\
 0 & -1 & 0& \\
 0 & 0 & 1 &
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}
=0
$$
$$
\begin{bmatrix}
x & y & 1
\end{bmatrix}
\begin{bmatrix}
\ 1 & \ 0 & \ 3 & \\
\ 0 & \ 1 & \ 3 & \\
\ 3 & \ 3 & \ 9 &
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}
=0
$$
Therefore, the equation can be expressed $(x+3)^2+(y+3)^2=9$ eventually.
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