General equation of a circle in 3D?

In summary, the general equation of a circle in 3D space can be described using the following identities: P = (X_{P},Y_{P},Z_{P}) = \mbox{General point on diameter of circle.} C = (X_{C},Y_{C},Z_{C}) = \mbox{Centre of the circle.} \bar{d} = \left(\begin{array}{cc}x_{d}\\y_{d}\\z_{d}\end{array}\right) = \mbox{Perpendicular to the plane of the circle.} r = \mbox{radius of circle}. These identities can be expanded to x_{d}(X_{C}-X
  • #1
Sandyscott
5
0
Hi,

I've been trying to deduce the general equation of a circle in 3D space, but without much luck.

Well, I'll jump in with what I've got so far.


[tex]P = (X_{P},Y_{P},Z_{P}) = \mbox{General point on diameter of circle circle.}[/tex]
[tex]C = (X_{C},Y_{C},Z_{C}) = \mbox{Centre of the circle.} [/tex]
[tex]\bar{d} = \left(\begin{array}{cc}x_{d}\\y_{d}\\z_{d}\end{array}\right) = \mbox{Perpendicular to the plane of the circle.} [/tex]
[tex]r = \mbox{radius of circle} [/tex]

These are the identities I've used to describe it:
[tex]\vec{PC}[/tex] is perperpendicular to [tex]\bar{d}[/tex] so
[tex]\vec{PC}.\bar{d}=0[/tex]

and,

[tex]|\vec{PC}| = r[/tex]

these expand (somewhat clumsily) to:
[tex]x_{d}(X_{C}-X_{P})+y_{d}(Y_{C}-Y_{P})+z_{d}(Z_{C}-Z_{P})=0 [/tex]
and
[tex]\sqrt{(X_{C}-X_{P})^2+(Y_{C}-Y_{P})^2+(Z_{C}-Z_{P})^2}=r[/tex]

Not sure what to do from here, or if I'm even barking up the right tree, any help would be much appreciated
 
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  • #2
Hi there Sandyscott and welcome to PF,

The easiest way is to define the 3D circle using parametric equations.
 
  • #3
And how do I do that?

Bearing in mind that in the problem I want to have the circle at a funny angle.
 
  • #4
Usually the equation of a circle in 3D is given by the locus of points that satisfies the eqn of some sphere and a plane which cuts off the circular cross section required.
What are the given parameters to determine the equation in your question ?
 
  • #5
This isn't for a formal maths question*, so I don't have any specifc boundaries. I think ideally I'd like to have parametric description of the curve in terms of one parameter, with these input constants: the centre of the circle, the perpendicular vector and the radius.


*I want to determine the shape of a piece of wood mouth of a duct (circular) and a rectangular hole in a model hovercraft I'm building (I have no intention of describing the piece of wood, I just want to "take measurements" from my mathematical model)
 
  • #6
Using vectors, generally if t is the parameter then and point P on the circle is given by;

[tex]P = R\cos(t) \vec{u} + R\sin(t) \;\;\vec{n}\times\vec{u} + c[/tex]

Where u is a unit vector from the centre of the circle to any point on the circumference; R is the radius; n is a unit vector perpendicular to the plane and c is the centre of the circle.
 
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  • #7
Thanks very much, that should be very useful.

Incidentally, is that derived from the intersection of a sphere and plane re: arunbg? (so i can have a go at understanding it myself ;-) )
 
  • #8
Sandyscott said:
Thanks very much, that should be very useful.

Incidentally, is that derived from the intersection of a sphere and plane re: arunbg? (so i can have a go at understanding it myself ;-) )

Yes it is derrived from the intersection of a sphere and a plane using vector geometry.
 
  • #9
Cool,

Thanks for the prompt responses, I've never had an answer as quickly in any forum.

Cheers
 
  • #10
Sandyscott said:
Cool,

Thanks for the prompt responses, I've never had an answer as quickly in any forum.

Cheers

My pleasure, if you need any help deriving the general form do not hesitate to post here again. :smile:
 
  • #11
Why does the equation of a 3d circle need a unit vector perpendicular to the plane and a unit vector from the center of the circle to any point on the circumference?
 
  • #12
play said:
Why does the equation of a 3d circle need a unit vector perpendicular to the plane and a unit vector from the center of the circle to any point on the circumference?
Because in three dimensions one needs both vectors to determine the vector that is tangential to the circumference at a given point and lies in the plane of the circle.
 
  • #13
Oh, thanks :)
 
  • #14
Hi,

How would one describe a circle in non-parametric form? I understand that a circle in 3d is the intersection of a sphere and a plane. Is it possible to have an equation describing the circle with only the following elements:
- coordinate of the center of the sphere (also a point on the plane), xc, yc, zc
- components of the vector normal to the plane, nx, ny, nz
- the radius of the sphere, r
- the locus of points of the circle which would of course be x, y, z

Thanks in advance.
 
  • #15
"Where u is a unit vector from the centre of the circle to any point on the circumference"

How exactly do you derive vector U? its more complicated than just Rcos(t),Rsin(t) because you have that third axis, right?
 
  • #16
Chris001 said:
"Where u is a unit vector from the centre of the circle to any point on the circumference"

How exactly do you derive vector U? its more complicated than just Rcos(t),Rsin(t) because you have that third axis, right?

I'm a little late to this thread, but as you have noticed, nobody has given an answer that contains only the three pieces of information posited by the original poster: The center, the radius, and a normal vector. Here's how to do that; the algebra gets messy in the general case but not bad for specific numbers.

Let [itex] N = \langle n_1,n_2,n_3\rangle[/itex] be the normal vector, [itex]\vec C = \langle a,b,c\rangle[/itex] be the position vector of the center, and r be the radius.

Let [itex]\vec u = N \times \vec i[/itex] and [itex]\vec v = N \times \vec u[/itex]. Normalize these two vectors to get [itex]\hat u[/itex] and [tex]\hat v[/itex] which are now orthogonal unit vectors in the plane of the circle. The parametric equation of the circle becomes:

[tex]\vec R(t) = \vec C + r\cos(t)\hat u + r\sin(t) \hat v[/tex]

[Edit] Additionally note that crossing N with i is to get a vector perpendicular to N. If N is parallel to i that won't work, but then just use j.
 
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  • #17
Could anyone please give the matrix form of the equation like:

[tex]X^T.A.X[/tex]​

with,

[tex]
X=\left(
\begin{array}{c}
x \\
y \\
z \\
1
\end{array}
\right)
[/tex]​

and [tex]A[/tex] the matrix.

Thanks in advance.
 
  • #18
Thanks for the general equation of a circle in 3D. I used it to find the parametric equation of an assumed-circular Earth orbit at a small inclination angle to the invariable (x,y) plane of the solar system. (The inclination angle varies up to 2 degrees with a ~100-kiloyear period. See http://www.muller.lbl.gov/pages/inclin3ma.html)

Now I want to use such a circle in an orbit surface integral and a line integral around the orbit with a Gaussian function (or other peaked function) weight as a function of the z variable to represent cosmic dust that accumulates in a narrow band in the invariable plane in both integrals. (I want both integrals because dust between the Sun and the Earth may have some climate effects and dust along the Earth's orbit may have some climate effects.

I could use some help in setting up the two integrals.

http://arts.bev.net/RoperLDavid/
 
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  • #19
I should have said up to 4 degrees instead of 2 degrees.
 

FAQ: General equation of a circle in 3D?

1. What is the general equation of a circle in 3D?

The general equation of a circle in 3D is (x-x0)2 + (y-y0)2 + (z-z0)2 = r2, where (x0, y0, z0) is the center of the circle and r is the radius.

2. How is the general equation of a circle in 3D derived?

The general equation of a circle in 3D can be derived by using the Pythagorean theorem and the equation of a circle in 2D (x-x0)2 + (y-y0)2 = r2. By adding the z-coordinate to both sides of the equation, we can represent the circle in 3D space.

3. What does each term in the general equation of a circle in 3D represent?

The term (x-x0)2 represents the squared distance between the x-coordinate of any point on the circle and the x-coordinate of the center of the circle. The same applies for (y-y0)2 and (z-z0)2 for the y and z coordinates respectively. The term r2 represents the squared radius of the circle.

4. Can the general equation of a circle in 3D be used to represent all circles in 3D space?

Yes, the general equation of a circle in 3D can be used to represent all circles in 3D space. By changing the values of (x0, y0, z0) and r, the equation can represent circles with different centers and radii.

5. How is the general equation of a circle in 3D used in real-world applications?

The general equation of a circle in 3D is used in various fields such as mathematics, physics, engineering, and computer graphics. It is used to represent and analyze circular objects in 3D space, such as spheres, orbits, and circular motion. It is also used in computer graphics to create 3D objects and animations.

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