How can I see that the space curve given by $\gamma(s) = (\frac 4 5 \cos s, 1 - \sin s, - \frac 3 5 \cos s)$ forms a circle in space ? Let $\gamma: \mathbb R \rightarrow \mathbb R^3$ be a space curve given by $\gamma(s) = (\frac 4 5 \cos s, 1 - \sin s, - \frac 3 5 \cos s)$.
How do I see that $\gamma$ has image in $\mathbb R^3$ that forms a circle ? Also how can I determine its radius and determine the plane it lies in ?
I've worked with circles in $\mathbb R^2$ given by equations of the form $(x-a)^2 + (y-b)^2 = r^2$. Can I utilize this knowledge here ?
I see the $x$-coordinate goes between $\pm \frac 4 5$, the $y$-coordinate between $0$ and $1$ and the $z$-coordinate between $\pm \frac 3 5$. Intuitively, this points towards a circle in space, but how can I determine this rigorously ?
 A: $\newcommand{\Vec}{\mathbf{u}}\newcommand{\Ctr}{\mathbf{c}}\newcommand{\Reals}{\mathbf{R}}$In general, if $\Vec_{1}$ and $\Vec_{2}$ are orthogonal vectors in $\Reals^{n}$ of equal length, say $\|\Vec_{i}\| = r > 0$, and if $\Ctr \in \Reals^{n}$ is arbitrary, then the "translated trigonometric linear combination"
$$
\gamma(s) = \Ctr + (\cos s)\Vec_{1} + (\sin s) \Vec_{2}
$$
parametrizes the circle of center $\Ctr$ and radius $r$ in the plane spanned by $\Vec_{1}$ and $\Vec_{2}$.
The proof is an easy, worthwhile exercise: Express $\|\gamma(s) - \Ctr\|^{2}$ as a dot product and expand using the assumptions about the $\Vec_{i}$. (Incidentally, if the $\Vec_{i}$ are linearly independent, $\gamma$ always parametrizes an ellipse.)
For the given $\gamma$, the center $\Ctr$ and vectors $\Vec_{i}$ are read out by inspection.
A: Since you seem to see what values the coordinates "go between", then you can already make an educated guess: what about the circle with center $\;(0,1,0)\;$?
$$x^2+(y-1)^2+z^2=\frac{16}{25}\cos^2x+\sin^2s+\frac9{25}\cos^2s=1$$
and there you go.
