I would use linear algebra for this. Turn the two points you have into a vector:
$(x_1-x_0)\hat{i}+(y_1-y_0)\hat{j} = \begin{pmatrix}x_1-x_0 \\ y_1-y_0\end{pmatrix}$
Next, multiply by the rotational matrix:
$\begin{pmatrix}\cos k & -\sin k \\ \sin k & \cos k\end{pmatrix}$
So, you have:
$\begin{align*}c & = \begin{pmatrix}x_0 \\ y_0\end{pmatrix} + \begin{pmatrix}\cos k & -\sin k \\ \sin k & \cos k\end{pmatrix}\begin{pmatrix}x_1 - x_0 \\ y_1-y_0\end{pmatrix} \\ & = \begin{pmatrix}x_0 + (x_1-x_0)\cos k-(y_1-y_0) \sin k \\ y_0 + (x_1-x_0) \sin k + (y_1-y_0)\cos k\end{pmatrix}\end{align*}$
That assumes that the distance from $(x_0, y_0)$ to C is the same as the distance from $(x_0, y_0)$ to $(x_1, y_1)$ which was probably what was intended but was not actually said in the first post.
Thanks, it is exactly what I needed. I was searchind and saw this method:
Cx=cos(θ)⋅(X1−X0)−sin(θ)⋅(Y1−Y0)+X0
Cy=sin(θ)⋅(X1−X0)+cos(θ)⋅(Y1−Y0)+Y1
The distance between P0 and P1 is the same of P0 and C.
Thanks one more time.