# Thread: Parameterising a certain circle

1. ## Parameterising a certain circle

Parameterise the circle that results from intersecting the unit sphere in $\mathbb{R}^3$ centered at the origin with the plane $[(x,y,z) \in : x = z ]$

I tried $x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1$
however the last equation is an ellipse????

2. Originally Posted by FGT12
Parameterise the circle that results from intersecting the unit sphere in $\mathbb{R}^3$ centered at the origin with the plane $[(x,y,z) \in : x = z ]$

I tried $x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1$
however the last equation is an ellipse????
the intersection is an ellipse and no circle!

3. You result is only the projection of the circle to the xy plane, which is really an ellipse.

4. so how would you parameterise the circle in $\mathbb{R}^3$

5. It's definitely going to be a circle, because really: how could a plane intersect a sphere and make an ellipse! You're right that it's absurd.

Have you tried drawing a sketch? Try visualizing the picture with the y axis pointing directly out of the paper, and that may give you some insight. I solved the problem by first coming up with an equation based on the sketch, then proving that it was right. So, like I said. Start with the picture.

6. ahaok is wrong, a sphere cut by a plane gives a circle, not an ellipse. xxp9 is right- a circle, projected onto a plane at an angle to a line perpendicular to the circle gives an ellipse. That is why you get the equation of an ellipse- you are projecting the circle onto the xy-plane, not parameterizing it.

Here, your sphere is the unit sphere, $x^2+ y^2+ z^2= 1$ and the plane is given by x= z. The "standard" parameterization of the unit sphere can be derived from spherical coordinates:
$x= \rho cos(\theta)sin(\phi)$
$y= \rho sin(\theta)sin(\phi)$
$z= \rho cos(\phi)$

by taking the radius variable, $\rho$ equal to 1:
$x= cos(\theta)sin(\phi)$
$y= sin(\theta)sin(\phi)$
$z= cos(\phi)$

The fact that z= x means that we must have $cos(\theta)sin(\phi)= cos(\phi)$ so that $cos(\theta)= \frac{cos(\phi)}{sin(\phi)}= cot(\phi)$.

So one way of getting the parameterization of the circle is to replace $\theta$ by $cos^{-1}(cot(\phi))$, reducing from two parameters to one:
$x= cos(cos^{-1}(cot(\phi)))sin(\phi)$
$y= sin(cos^{-1}(cot(\phi))sin(\phi)$
$z= cos(\phi)$

Of course, $cos(cos^{-1}(cot(\phi))= cot(\phi)$. And, we can use $sin(\theta)= \sqrt{1- cos^2(\theta)}$ to say that
$sin(cos^{-1}(cot(\phi))= \sqrt{1- (cos(cos^{-1}(cot(\phi)}= \sqrt{1- cot^2(\phi)}$ and so have

$x= cot(\phi)sin(\phi)= cos(\phi)$
$y= \sqrt{1- cot^2(\phi)}sin(phi)$
$z= cos(\phi)$.

7. Originally Posted by FGT12
Parameterise the circle that results from intersecting the unit sphere in $\mathbb{R}^3$ centered at the origin with the plane $[(x,y,z) \in : x = z ]$

I tried $x^2 + y^2 + z^2 = 1 \Rightarrow 2x^2 + y^2 = 1$
however the last equation is an ellipse????
The vector form for the plane is

$\vec{r}(s,t)=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}t+\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}s$

This gives that $x=z=t$ Using the equation of the sphere gives

$t^2+y^2+t^2=1 \implies y=\pm\sqrt{1-2t^2}$

So a parametric representation of the circle is
$x=t,\quad y=\pm\sqrt{1-2t^2},\quad z=t$

Since the plus or minus is really messy it can be eliminated using trigonometry.

If we define $\sin(\alpha)=\sqrt{1-2t^2}$ this gives
$\cos(\alpha)=\frac{t\sqrt{2}}{1} \implies t=\frac{1}{\sqrt{2}}\cos(\alpha)$

Using this we get

$\displaystyle \vec{r}(\alpha)=\left( \frac{1}{\sqrt{2}}\cos(\alpha)\right)\vec{i}+\sin( \alpha)\vec{j}+\left( \frac{1}{\sqrt{2}}\cos(\alpha)\right)\vec{k}$

Edit: too slow