Fin the Rotation $\alpha$ Centered in $(2,1) \in \mathbb{R^2}$. Let $\alpha \in [0, 2\pi]$. Write the rotation $\alpha$ centered in $(2,1) \in \mathbb{R^2}$ under the form Q(x) + v
attempt at finding the solution
Let x =$\left[ \begin{array}{ccc}
x_1 \\
x_2 \\
\end{array} \right]$
The rotation matrix is $\left[ \begin{array}{ccc}
cos\alpha \ -sin\alpha \\
sin\alpha   \   cos\alpha \\ 
\end{array} \right]$
Therefore the new coordinates of a point $x'$ after rotation is:
$x' =\left[ \begin{array}{ccc}
x'_1 \\
x'_2 \\
\end{array} \right]$ =$ \left[ \begin{array}{ccc}
cos\alpha \ -sin\alpha \\
sin\alpha   \   cos\alpha \\ 
\end{array} \right]\left[ \begin{array}{ccc}
x_1 \\
x_2 \\
\end{array} \right]$ =$\left[ \begin{array}{ccc}
x_1cos\alpha - x_2sin\alpha \\
x_1sin\alpha + x_2cos\alpha  \\
\end{array} \right] $
But the rotation is centered in $[0, 2π)$ Therefore the rotation is $\left[ \begin{array}{ccc}
x_1cos\alpha - x_2sin\alpha \\
x_1sin\alpha + x_2cos\alpha  \\
\end{array} \right] +\left[ \begin{array}{ccc}
2 \\
1 \\
\end{array} \right] =\left[ \begin{array}{ccc}
cos\alpha \ -sin\alpha \\
sin\alpha   \   cos\alpha \\ 
\end{array} \right]\left[ \begin{array}{ccc}
x_1 \\
x_2 \\
\end{array} \right] + \left[ \begin{array}{ccc}
2 \\
1 \\
\end{array} \right]$
end of attempt
Is that the right approach to the question?
 A: You are wrong because the rotation matrix 
$$
R=
\begin{bmatrix}
\cos \alpha& -\sin \alpha\\
\sin \alpha&  \cos \alpha
\end{bmatrix}
$$
represents a rotation centered at $(0,0)$, so the first step is to translate the origin to the center of rotation $[2,1]^T$ . This is done by the translation:
$$
\vec x=\begin{bmatrix} x\\y\end{bmatrix} \rightarrow \vec x'=\begin{bmatrix} x-2\\y-1\end{bmatrix}=\begin{bmatrix} x\\y\end{bmatrix}-\begin{bmatrix} 2\\1\end{bmatrix}=\vec x - \vec t
$$
Where $\vec t$ is the center of rotation.
Now we can do the rotation:
$$
R \vec x'= \begin{bmatrix}
\cos \alpha& -\sin \alpha\\
\sin \alpha&  \cos \alpha
\end{bmatrix}\begin{bmatrix} x-2\\y-1\end{bmatrix}=
\begin{bmatrix} (x-2)\cos \alpha-(y-1)\sin \alpha\\(x-2)\sin \alpha+(y-1)\cos \alpha\end{bmatrix}
$$
Finally we have to return to the old coordinate system with a translation $\vec t$, so the final result is:
$$
R^*\vec x =R\vec x' +\vec t
$$
that you can write as:
$$
R^*\vec x=R(\vec x-\vec t)+\vec t=R\vec x +(I-R)\vec t
$$
