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
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.