Radius and center of a circle in $\mathbb{R}^3$.

Question

If there is a circle in $\mathbb{R}^3$, for example: $(\frac{\cos\phi}{\sqrt{2}-\sin\phi},\frac{\sin\phi}{\sqrt{2}-\sin\phi},\frac{\cos\phi}{\sqrt{2}-\sin\phi})$ where $\phi \in [0,2\pi]$ (Which looks like a circle, according to my plot in wolfram mathematica) How can I calculate its radius and center? Calculating the squares of the coordinates does not give constans. Is there any general method?

Answer 1

HINT: Call $\gamma(t)$ the parametrization of your curve. Then set $P=(x_0,y_0,z_0)$ and compute the difference $$ \|\gamma(t)-P\|^2\;, $$ trying to find the coordinate for $P$ such that the above quantity results constant wrt $t$.

Answer 2

We try to find two algebraic dependencies between the coordinates of the points on the curve. The first is obvious, $x=z$, the second with some help from WA : $$x^2 + (y-1)^2 + z^2 = 2$$

Another approach: solve and get $$\cos \phi = \frac{\sqrt{2}\, x}{y+1}\\ \sin \phi= \frac{\sqrt{2}\, y}{y+1}$$ so $2 x^2 + 2 y^2 =(y+1)^2$, and using $x=z$ we get the same equation as above. Note that every point on the circle except $(0,-1,0)$ is covered, and exactly once.