Showing that a given curve is a circumference

Question

Consider the curve $\beta(s)=(\frac{4}{5} \cos(s),1-\sin(s), \frac{-3}{5} \cos(s))$. Prove that the trace of $\beta$ is a circumference.

My approach:

$(*)$ If there exists a point $p \in \mathbb{R}^3$ such that for any point $\beta(s)$ the distance between $\beta(s)$ and $p$ is constant, then the trace of $\beta(s)$ is either a circumference or a sphere (is this true?).

I have already proved that $\beta$ is planar, showing that its torsion is constantly zero, but I have failed to prove $(*).$ Any hints or different approaches?

Answer 1

$$ \left(\frac{4}{5} \cos s,1-\sin s, \frac{-3}{5} \cos s\right) \tag 1 $$

The set of all points of the form $\big(x,y,(-3/4)x\big)$ for $x,y\in\mathbb R$ is a plane.

Let $x = \dfrac 4 5 \cos s$ and let $y= 1-\sin s,$ and we see that as $s$ varies, all points of the curve $(1)$ lie within that plane.

As $s$ varies, the average value of $(1)$ is $(0,1,0),$ so if $(1)$ is a circle then that must be the center. Show that the distance from a point of the form $(1)$ to $(0,1,0)$ does not depend on $s.$

Answer 2

Model the circle as $p+ (\cos t) v_1+(\sin t) v_2$, where $v_1 \bot v_2$ and $\|v_1\|= \|v_2\|$.

Once you compute $p$, it is not hard to guess a suitable $v_1, v_2$ for the given $\beta$.

Note that the period is $2 \pi$.

Compute $p={1 \over 2} (\beta(0)+ \beta(\pi))= (0,1,0)$. We can read off $v_1= ({4 \over 4}, 0, - {3 \over 5})$ and $v_2 = (0,-1,0)$.