# Find The Equation of a Circle in R^3

• CrankFan
In summary: The question is asking for the equation of the great circle that passes through the point (1 / sqrt(6), -1 / sqrt(6), 2 / sqrt(6)) not associated with it. The family of great circles that pass through (1 / sqrt(6), -1 / sqrt(6), 2 / sqrt(6)) can be found by rotating about the point (1 / sqrt(6), -1 / sqrt(6), 2 / sqrt(6)).
CrankFan
Hi.

This question is from the book Geometry by Brannam, Esplen, Gray and it comes from the chapter 7, section 1 review problems which cover introduction to spherical geometry.

Determine the equation of the great circle that passes through the point (1 / sqrt(6), -1 / sqrt(6), 2 / sqrt(6))

This is probably a dumb question but I'm not even sure what the general equation for a circle in R^3 looks like, if I knew that then I think I could answer the question. I'm thinking (...without much justification...) that perhaps the equation of the circle in R^3 is like that of a plane in R^3? Where the coefficients of the x, y and z variables in the standard form, are the components of the normal vector ..? In other words I think I need to incorporate the normal to uniquely identify the specific circle of radius 1 which is centered at the origin because there are so many, but I'm not really sure how to do that and haven't had much success so far, and am a bit frustrated now

Thanks.

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If you can find a sphere passing through those points, and the plane passing through those points, then the intersection is the circle

First, you need to remember that you cannot write a one dimensional object, such as a circle, in R3 as a single equation! You can write in terms of two equations (the intersection of two surfaces) or as 3 parametric equations in one parameter.

The question, as you give it, is not "well defined". You are asked not just for a circle but for the great circle. A great circle is a circle on a sphere having the same center as the sphere. In addition, a circle or sphere is not determined by one point.

Since the given point has distance 1 from the origin, I'm guessing that you are asked to find equations for the great circle on the unit sphere (that is, the sphere with center at (0,0,0) and radius 1) that passes through
$$\left(\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right)$$
Even so, there exist an infinite number of such great circles. You would need another point on the circle to define this exactly.

HallsofIvy said:
First, you need to remember that you cannot write a one dimensional object, such as a circle, in R3 as a single equation!

I had a suspicion that that is the case but didn't want to jump to any conclusions. Yes. We're talking about the unit sphere: S^2, sorry for not making that clearer.

I scanned the page from the book and posted it here so that people can see the exact question:

http://img528.imageshack.us/img528/3149/brennansaa3.th.jpg

The question is 1(a) of the 7.1 review problems.

Now that I think about it you're right. Infinitely many great circles pass through a given point. Seems like the more math I learn the more my intuition sucks.

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I'm not certain about what the author of the book wants. It's mentioned in section (7.1) that the great circle associated with the point P means the unique great circle of points $$\frac{\pi}{2}$$ distant from P.

Determine the equation of the great circle associated with the point $$\left(\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right)$$ Then I think I could use the fact that OP is the normal to the plane that the great circle lies on and use the equation of the plane:

$$\frac{1}{\sqrt{6}}x -\frac{1}{\sqrt{6}}y + \frac{2}{\sqrt{6}}z = 0$$ and the eqn of the the unit sphere $$x^2 + y^2 + z^2 - 1 = 0$$ to get the equation of the great circle:

$$\left(x^2 - \frac{1}{\sqrt{6}}x \right) + \left(y^2 + \frac{1}{\sqrt{6}}y \right) + \left(z^2 - \frac{2}{\sqrt{6}}z \right) - 1 = 0$$

Then do the algebra.

But the question asks about the equation of a great circle that passes through the point $$P\left(\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right)$$, not associated with it, and as Halls pointed out there are a lot of those. I think that any great circle that passes through P, rotated about OP gives the family of great circles that pass through P. Maybe the authors wanted something like that? It's not clear to me how to get the rotation matrix for rotation about OP but does it seem possible that that is what the author was asking, or maybe it seems like the altered version of the problem I gave above was more likely? Section 7.1 only touches upon rotations as an intuitive concept, section 7.2 introduces elementary rotations and discusses finding a 3 x 3 matrix that corresponds to a rotation that maps a given point P1 to P2.

Thanks Halls & Office Shredder.

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## 1. What is the general equation of a circle in R^3?

The general equation of a circle in R^3 is (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2, where (a, b, c) is the center of the circle and r is the radius.

## 2. How do you find the center and radius of a circle in R^3?

To find the center and radius of a circle in R^3, you need three points on the circle. You can then use the distance formula to find the distance between these points, which will be equal to the radius. The center of the circle can be found by finding the midpoint of any two of the three points.

## 3. What is the difference between a circle in R^2 and a circle in R^3?

A circle in R^2 is a two-dimensional object, while a circle in R^3 is a three-dimensional object. This means that a circle in R^2 can be represented by a single equation, while a circle in R^3 requires three equations to define it.

## 4. Can you have a circle in R^3 with a negative radius?

Yes, it is possible to have a circle in R^3 with a negative radius. This would result in a circle that is inside out, with the center of the circle on the outside and the outer edge on the inside.

## 5. How do you graph a circle in R^3?

To graph a circle in R^3, you can plot the center point and then use the radius to plot points along the circle. You can also use a 3D graphing calculator or software to plot the circle in three-dimensional space.

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