The motion of the particle satisfies $\textbf{v} = \textbf{c}\times \textbf{r}$

Question

Why is the path is contained in a circle that lies in a plane perpendicular to $\textbf{c}$ with centre on a line through the origin in the direction of $\textbf{c}$

Answer 1

$$\textbf{v}.\textbf{r}=(\textbf{c}\times\textbf{r}).\textbf{r}=0\tag{1}$$ and $$\textbf{v}.\textbf{c}=0\tag{2}$$ Therefore $\textbf{v}\perp\textbf{r}~~\forall t$ and $\textbf{v}\perp\textbf{c}$. $$\frac{d}{dt}(\textbf{r}.\textbf{c})=\textbf{v}.\textbf{c}=0\implies \textbf{r}.\textbf{c}=constant\tag{3}$$ which is the equation of a plane orthogonal to $\textbf{c}$. $$\frac{d}{dt}\left(\textbf{r}^2\right)=2\textbf{v}.\textbf{r}=0\implies ||\textbf{r}||=k=constant\tag{4}$$ which is the equation of a circle of center the origin lying in the plane $(3)$.