Find the radius, center and plane of a circle

Question

I have the parametrization

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

I now, that its curvature is constant, and torsion is zero, hence the parametrization is a circle.

How can I now find the centre of the circle, the radius and the plane in which it lies?

Answer 1

Notice that all components have a periodicity of $2\pi$. Choose three points, equally spaced. You can choose any three s values, but $0, 2\pi/3, 4\pi/3$ would make calculations easier. The circumcenter of these points is the center of the circle. The distance from the center to any of these points is the radius. And from three points you can get the equation of the plane.

Answer 2

HINT

For the center and radius we can choose two opposite points $P$ and $Q$ (at $0$ and $\pi$ for example) and calculate medium point $O=\frac{P+Q}{2}$ that is the center $O$ and distance $OP$ that is the radius.

For the plane you can calculate velocity i vector $\vec v$ at $P$ and obtain the normal to the plane by cross product $\vec n=\vec v\times \vec {OP}$.

Answer 3

Hint: You have

$$ x = \frac45 \cos s, \quad y = 1 - \sin s, \quad z = -\frac35 \cos s $$

Therefore $$ x^2 + z^2 + (1-y)^2 = \cos^2 s + \sin^2 s = 1 $$

This is a sphere with center $(0,1,0)$ and radius $1$. The parametrization is a big circle of this sphere, so it has to have the same center and radius.

Also, observe that $$ 3x + 4z = 0 $$ since both $x$ and $z$ are proportional to $\cos s$.

This happens to be the equation of a plane. The intersection of this plane and the above sphere results in your parametrization.