# Thread: Equation of a circle in 3d

1. ## Equation of a circle in 3d

Dear all,
I am trying to work out the equation for a circle in 3d.

Circle with centre (a,b,c)
Perpendicular to vector [a,b,c]

In pursuit, I have found that this circle is where the plane
$ax+by+cz=a^2+b^2+c^2$
intersects the sphere
$(x-a)^2+(y-b)^2+(z-c)^2=r^2$

...but I'm not sure where to go from there, or indeed whether that is even useful.

All suggestions welcome!
Many thanks

2. One possible way:
Solve for z = f(x,y) from the first equation.
Substitute into the second equation to eliminate z.
Now use this to find y = g(x).

Use x = t, y = g(t), and z = f(t,g(t)).
This gives at least one part of the cricle in parametric form.

3. Originally Posted by sevenquid
Dear all,
I am trying to work out the equation for a circle in 3d.

Circle with centre (a,b,c)
Perpendicular to vector [a,b,c]

In pursuit, I have found that this circle is where the plane
$ax+by+cz=a^2+b^2+c^2$
intersects the sphere
$(x-a)^2+(y-b)^2+(z-c)^2=r^2$

...but I'm not sure where to go from there, or indeed whether that is even useful.

All suggestions welcome!
Many thanks
hai,
sphere is having the centre as (a,b,c)
and this point satisfies the plane equation, so plane is passing through the centre of the sphere.

radius of the circle will be same(r)
since circle is in two dimension centre can be (a,b) (b,c) 0r (a,c).
that enough we can find equation of the circle.

4. snowtea I followed your suggestion: what I get home later I'll see if Autograph 3.3 can handle it.
I am surprised (and concerned) by its complexity

5. My picture link has broken: I'll try again: Twitpic - Share photos on Twitter
That's my result. Maple 12 seems to handle it, but isn't rendering it too well, or there's a bug in what I've programmed it to display.

An equation for the circle in spherical polar coordinates would probably be a lot simpler and easier to work with.
I'd really appreciate some help with coming up with the equation in spherical polar coords.

Bearing in mind this isn't an entirely skew circle - it is perpendicular to the radial vector - the problem doesn't seem like it should be that difficult

6. You know the equation for a unit circle in the xy plane (z-axis normal) is:
$x = cos(t), y = sin(t), z = 0$

Moving the center and scaling the radius is the easy part.
The main difficulty is changing plane of the circle. Rotating from a z-axis normal to one with normal vector (a,b,c).

So we want to find the rotation transformation that takes the vector (0,0,1) to (a,b,c).
There are an infinite number of different rotations.
The most intuitive one would be to rotate about the axis $(0,0,1)\times (a,b,c)$ by $\theta = cos^{-1}(\frac{c}{\sqrt{a^2 + b^2 + c^2}})$.
There is a well known formula to find the rotation matrix given axis and angle: Rotation matrix - Wikipedia, the free encyclopedia

Once you find your rotation matrix $R$. You new parameteric equations are $x'(t), y'(t), z'(t)$ given by:
$(x'(t), y'(t), z'(t)) = R(x(t),y(t),z(t)) = R(cos(t),sin(t),0)$

7. Originally Posted by snowtea
You know the equation for a unit circle in the xy plane (z-axis normal) is:
$x = cos(t), y = sin(t), z = 0$

Moving the center and scaling the radius is the easy part.
The main difficulty is changing plane of the circle. Rotating from a z-axis normal to one with normal vector (a,b,c).

So we want to find the rotation transformation that takes the vector (0,0,1) to (a,b,c).
There are an infinite number of different rotations.
The most intuitive one would be to rotate about the axis $(0,0,1)\times (a,b,c)$ by $\theta = cos^{-1}(\frac{c}{\sqrt{a^2 + b^2 + c^2}})$.
There is a well known formula to find the rotation matrix given axis and angle: Rotation matrix - Wikipedia, the free encyclopedia

Once you find your rotation matrix $R$. You new parameteric equations are $x'(t), y'(t), z'(t)$ given by:
$(x'(t), y'(t), z'(t)) = R(x(t),y(t),z(t)) = R(cos(t),sin(t),0)$
This has helped me a great deal. Much appreciated

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