# A circle on the plane [duplicate]

Possible Duplicate:
Parametric Equation of a Circle in 3D Space?

I know that, for example, if a circle is on a plane with counter-clockwise orientation, and with center $(a,b)$ and radius $R$, it has parametrization

$$r(t)=(a + R \cos{t};b + R \sin{t}) \quad 0 \leq t \leq 2\pi$$

and with clockwise orientation

$$r(t)=(a + R \sin{t},b + R \cos{t}).$$

Also, I know of forms of circle parametrization if it lies on horizontal plane $z=c$ and center $O(a,b,c)$, or if it is located on the plane $x=c$, I know how to parametrize the circle in this case. I am interested in what happens if the circle does not lie in any plane parallel to the coordinate planes?

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"i am interested what if f the circle does not lie in any plane parallel to the coordinate planes" - then you use an appropriate rotation matrix on the parametric equations for a circle. –  Ｊ. Ｍ. Jul 27 '12 at 13:36
what kind of appropriate matrix? –  dato datuashvili Jul 27 '12 at 13:39
Do you know how rotation matrices are built up? –  Ｊ. Ｍ. Jul 27 '12 at 13:43
generally no,i now some kind of rotation matrices,but how they are built up,i dont know –  dato datuashvili Jul 27 '12 at 13:44
I did no such thing; why did you automatically assume that I was your downvoter? In any event, I am advicing you to read up a bit on rotation matrices first, and then we can talk. –  Ｊ. Ｍ. Jul 27 '12 at 13:49

## marked as duplicate by Ｊ. Ｍ., t.b., Chris Eagle, Nate Eldredge, Zev ChonolesJul 28 '12 at 15:13

Let $\mathbf{u},\mathbf{v}$ be any two orthonormal vectors in $\mathbb{R}^n$, let $\mathbf{a} \in \mathbb{R}^n$ and let $R > 0$ be a positive real number. Then the circle of radius $R$ with centre $\mathbf{a}$ lying in the plane through $\mathbf{a}$ which is parallel to $\mathbf{u}$ and $\mathbf{v}$ is given by

$$\mathbf{r}(t) = \mathbf{a} + (R\cos t)\mathbf{u} + (R\sin t)\mathbf{v}$$

where $\mathbf{r}(t)$ denotes the locus of the points on the circle.

So given a plane $\Pi \subseteq \mathbb{R}^3$, calculate $\mathbf{u}$ and $\mathbf{v}$ and substitute into the above.

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$u$ and $v$ we can calculate using this parametrizations,which i have written above or? –  dato datuashvili Jul 27 '12 at 14:04
@dato: I don't really understand what you're asking. You can calculate $\mathbf{u}$ and $\mathbf{v}$ by looking at the specification of the plane you want your circle to be in. There are a number of ways of specifying a plane, which I'm sure you've come across already. –  Clive Newstead Jul 27 '12 at 14:06
aa ok ok thanks very much,just one question $a$ in this case is a scalar right?i dont want to mix,or it is vector? –  dato datuashvili Jul 27 '12 at 14:08
It's a vector. It couldn't possibly be a scalar, because you can't add a scalar to a vector. Here, $\mathbf{a}$ is the centre of the circle, so you could write, say, $\mathbf{a} = (a,b,c)$ if you wanted to. (At least, you could when working in 3 dimensions.) –  Clive Newstead Jul 27 '12 at 14:13
thanks very much for answer –  dato datuashvili Jul 27 '12 at 14:14
Assume your circle $\gamma$ has center ${\bf p}\in{\mathbb R}^3$. The plane $\Pi$ in which the circle lies has to be given somehow. If the plane is given in the form ${\bf n}\cdot{\bf x}=\rho$, where $\rho={\bf n}\cdot {\bf p}$, then we first have to provide ourselves with two mutually orthogonal unit vectors ${\bf e}_i$ which "span $\Pi$" or, to be exact, are orthogonal to ${\bf n}$. The first of these vectors can be obtained by normalizing the vector ${\bf e}_1':=(-n_2,n_1,0)$ (which is orthogonal to ${\bf n}$). Normalizing means replacing the vector ${\bf e}_1'$ by the unit vector $${\bf e}_1:={{\bf e}_1'\over\bigl|{\bf e}_1'\bigr|}$$ which points in the same direction. The second vector ${\bf e}_2$ is then obtained by normalizing the vector ${\bf e}_2':={\bf n}\times{\bf e}_1$.
After these preparatory steps it is easy to write down a parametric representation of $\gamma$: $$\gamma:\quad t\mapsto{\bf p}+R\cos(t){\bf e}_1+R\sin(t){\bf e}_2\qquad(0\leq t\leq 2\pi)\ .$$
sorry normalize vector $e_1=(-n_2,n_1,0)$ what does mean?which vector we are normalizing –  dato datuashvili Jul 27 '12 at 14:23