Equation of rotational curves around a point

Question

Let the figure below where a curve $f(x,y) = 0$ is known (the black one). I would like to rotate this curve around a point $P$ in both counterwise and counterclockwise directions by an angle $\theta$.

What are the equations of the blue and red curves?

My thoughts:

The tangent at any point is rotated by $\theta$, but it is not clear to me how to find these equations. Moreover, $f(x,y) = 0$ cannot be necessarily written in the form $y = g(x)$, so that using tangents at each point can become complicated!

Thanks for any hint!

Answer 1

Note $(p_x, p_y)$ the coordinates of $P$. The rotation $R_\theta$ of angle $\theta$ around the point $P$ is such that the coordinates of a point $M=(x,y)$ will be transformed into a point of coordinates $M^\prime=(x^\prime,y^\prime)$ with following relations: $$\begin{pmatrix} x - p_x\\y - p_y\end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \\-\sin \theta & \cos \theta\end{pmatrix}\begin{pmatrix} x^\prime - p_x\\y^\prime - p_y\end{pmatrix}$$

Based on those matrix relations, you can replace $(x,y)$ in the equation $f(x,y)=0$ to get the equation of the rotated curve. The other one is obtained replacing $\theta$ with $-\theta$.

Answer 2

Hint: Consider the set$$\left\{(x,y)\in\mathbb{R}^2\,\middle|\,f\bigl(x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta)\bigr)=0\right\}.$$