Showing that a Unit Speed Curve is a Circle. In my recent differential geometry tutorial, we were given the question:

Given the unit speed curve,
$$\boldsymbol{r}(s)=\left(\frac{4}{5}\cos(s),1-\sin(s),-\frac{3}{5}\cos(s)\right)$$
show that this represents a circle with centre $(0,1,0)$ with radius 1.

My first intuitive thought is to simply find the distance between $(0,1,0)$ and $\boldsymbol{r}(s)$ and show that it is 1 for all $s$ - this is the definition of a circle, correct? However, my tutor advised that we had to look into the torsion of the curve and use the fact that it is 0.
Any help/advice would be greatly appreciated. Thank you in advanced.
 A: Notice that in three dimension, a curve of constant distant to a fixed point is a curve on the sphere. So not only you shall show what you mentioned, but that torsion vanishes to ensure the curve is restricted to a plane, hence a circle in 2 dimensional subspace.
A: By looking at x,y and z coordinates, notice that the curve is intersection of an elliptic cylinder and plane through y-axis at a certain angle to get equal axes for the ellipse.. as circle.Directly use Frenet-Serret to find that torsion vanishes, curvature is constant.
A: With the rotation matrix around axis $Oy$,
$$R_\theta=\left(\begin{matrix}
\cos \theta & 0 & -\sin \theta \\
0 & 1 & 0 \\
\sin \theta & 0 & \cos \theta \\
\end{matrix}\right)$$
And
$$r=\left(\begin{matrix}
\frac45\cos s\\
1-\sin s\\
-\frac35\cos s
\end{matrix}\right)$$
Then
$$R_\theta \cdot r=\left(\begin{matrix}
\frac{(3\sin \theta + 4\cos \theta)\cos s}5\\
1-\sin s \\
\frac{(4\sin \theta - 3\cos\theta)\cos s}5 \\
\end{matrix}\right)$$
Thus, with $\theta=\mathrm{Arctan} \frac34$, the third component vanishes, and (you've got a $3-4-5$ triangle):
$$\cos \theta=\frac45, \ \sin\theta=\frac35$$
So
$$R_\theta \cdot r=\left(\begin{matrix}
\frac{(3\sin \theta + 4\cos \theta)\cos s}5\\
1-\sin s \\
0 \\
\end{matrix}\right)=\left(\begin{matrix}
\cos s\\
1-\sin s \\
0 \\
\end{matrix}\right)$$
So the rotated curve has a simpler equation.
