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 ?

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.
