Rotating $(3,3)$ by $45^\circ$ about center $(2,1)$ 
With a two dimensional surface, if we take $(2,1)$ as the center point and consider a transformation with a rotation axis around $45^\circ$, then point $(3,3)$ is transformed into point $(?,?)$

I am a bit stumped on how to do a $45^\circ$ rotation. I'd prefer answers that steer clear of using a rotation matrix.

While the rotation matrix that rotates $(x, y)$ by angle $\theta$ counterclockwise around a center $(c_x, c_y)$ may look scary,
$$\left[\begin{matrix} x^\prime \\ y^\prime \end{matrix}\right] = 
\left[\begin{matrix} c_x \\ c_y \end{matrix}\right] + \left[\begin{matrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta \\
\end{matrix}\right] \left[ \begin{matrix} x - c_x \\ y - c_y \end{matrix} \right]$$
it is only a way to write
$$\left\lbrace ~ \begin{align}
x^\prime &= c_x + (\cos\theta)(x - c_x) - (\sin\theta)(y - c_y) \\
y^\prime &= c_y + (\sin\theta)(x - c_x) + (\cos\theta)(y - c_y) \\
\end{align} \right.$$
where $(x^\prime, y^\prime)$ are the coordinates of the rotated point.

When $\theta = 45°$, $\cos\theta = \sqrt{1/2}$ and $\sin\theta = \sqrt{1/2}$.  So, to rotate point $(3, 3)$ by $45°$ around point $(2,1)$:
$$\left\lbrace ~ \begin{align}
x &= 2 + \sqrt{\frac{1}{2}}(3 - 2) - \sqrt{\frac{1}{2}}(3 - 1) \\
y &= 1 + \sqrt{\frac{1}{2}}(3 - 2) + \sqrt{\frac{1}{2}}(3 - 1) \\
\end{align} \right.$$
which is approximately $(1.293, 3.121)$.
