Show that the motion is a circular path Question:
The position $\mathbf r$ of a particle satisfies
$$ \ddot {\mathbf r} = f\dot {\mathbf r} \times \mathbf k$$
where $f$ is a known constant and $\mathbf k$ is the unit vector in the $z$ direction.
Given that $\dot {\mathbf r} \cdot \mathbf k \equiv 0$, show that the path of the particle is a circle.

Attempt:
Firstly, there are a few quantities that I know are constant:
$$\frac{d}{dt}\big(|\dot{\mathbf r}|^2 \big) = 2\ddot {\mathbf r} \cdot \dot {\mathbf r} = 2f(\dot{\mathbf r} \times \mathbf k) \cdot \dot{\mathbf r} = 0 \implies |\dot{\mathbf r}|^2 \equiv \text{constant}$$
$$\frac{d}{dt}\big(\mathbf r \cdot \mathbf k\big) = \dot{\mathbf r} \cdot \mathbf k =0 \implies {\mathbf r} \cdot \mathbf k \equiv \text{constant}$$
Moreover, I noticed that
$$\frac{d^2\dot{\mathbf r}}{dt^2} = f^2\dot{\mathbf r}$$
and solving this gives
$$\dot{\mathbf r}(t) = \mathbf c_1 \cos(ft)+\mathbf c_2 \sin (ft)$$
but I am unsure how to deduce from this that $\mathbf r$ traces out a circle.

Let $\mathbf u = \mathbf r + \frac1f (\dot{\mathbf r} \times \mathbf k)$, we have
$$\dot{\mathbf u} = \dot{\mathbf r} + \frac1f(\ddot{\mathbf r} \times \mathbf k)
= \dot{\mathbf r} + ((\dot{\mathbf r} \times \mathbf k) \times \mathbf k)
= \dot{\mathbf r} + ((\dot{\mathbf r}\cdot \mathbf k) \mathbf k - \dot{\mathbf r}) = \mathbf 0$$
This means $\mathbf u$ is a constant. Notice
$$\frac{d}{dt}|\mathbf r - \mathbf u|^2
= 2(\mathbf r-\mathbf u)\cdot \dot{\mathbf r} = -\frac{2}{f} (\dot{\mathbf r} \times \mathbf k)\cdot \dot{\mathbf r} = \mathbf 0$$
The expression $|\mathbf r-\mathbf u|^2 = R^2$ is a constant. The particle is moving on a sphere
centered at $\mathbf u$ with radius $R$. Together with what you know $\mathbf r \cdot \mathbf k = $ constant.
The particle is moving along the intersection of a sphere and a plane, i.e. a circle.
