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Firstly I thought that it might be 2. Having point A as the center. The vector between A and another point B is perpendicular to the plane of the circle and its length determines the radius of the circle. But then I thought this might define a disk. Thoughts ?

Kendall
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  • Any circle with a diameter at least the distance between two points can be made to pass through the two points. – Richard Ambler Dec 20 '16 at 09:53
  • In a plane, you need either three points on the circle, or its center and one additional point on the circle (in this case, only 2). In 3-D you always need 3. Even if you have the center and one point on the circle, the circle it is not uniquely defined since the plane in which the circle lies is not unique. – Florian Dec 20 '16 at 10:11
  • @RichardAmbler The OP’s second point isn’t on the circle. It’s used to define both a normal to the plane of the circle and its radius. – amd Dec 20 '16 at 20:15
  • @Florian The OP’s proposed second point is not meant to be on the circle. It defines the circle’s radius and plane. – amd Dec 20 '16 at 20:16
  • @amd you are right, I misread the description. My apologies. In this case, the OP is right, you could encode all the required information into two points: one being the circle center and thus in the plane, the difference between the two defining the direction of the normal vector (and thus the plane) and its length defining the radius. It's not an intuitive way but it's possible. – Florian Dec 20 '16 at 22:47

4 Answers4

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Three points are needed. They define uniquely a plane where a circle lies and the circle as well.

Widawensen
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  • Through 2 points in 3-d space it can can be generated infinite number of planes. (They have common straight line). Through 2 points on the plane also it can be generated infinite number of circles. So two points are not sufficient. – Widawensen Dec 20 '16 at 09:55
  • Don't need to define a unique circle. Just at least 1 – Kendall Dec 20 '16 at 10:18
  • @user29171 Three points give you precisely 1 circle ( to be honest they should not be collinear). Two points allow to generate infinite number of circles. – Widawensen Dec 20 '16 at 10:31
  • Doesn't the vector ensure that it makes only 1 circle though, since there is only one plane normal to this vector ? – Kendall Dec 20 '16 at 10:41
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    @user29171 ..and you want one endpoint of this vector to be the center of a circle and the second to define the radius and plane.. In this case you have all parameters to construct the circle. – Widawensen Dec 20 '16 at 11:12
  • This answer doesn’t address the proposal in OP’s question, which doesn’t involve points on the circle. – amd Dec 20 '16 at 20:20
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A circle is the boundary of a disk, so if you can characterize a disk uniquely, then you can also characterize a circle uniquely. Your considerations are correct.

Dominik
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It depends how you use the points to define the circle.

If you want the circle to pass through the points, then three points are needed.

But there is another way to define a circle from two given points $P$ and $Q$: you use one point, $P$, as the center of the circle, and you use the other point $Q$ to define the plane containing the circle and its radius. Specifically, the plane normal is in the direction of the vector $Q-P$, and the circle radius is $\|Q - P\|$.

bubba
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I know this is a necro, but the answer is that you only need a single real number to uniquely define a circle in any finite dimensional space.

First, note that 3 points define a circle. So, all we need is a surjective function $f: \mathbb{R}—>\mathbb{R^n} \times \mathbb{R^n} \times \mathbb{R^n}$ (i.e. $\mathbb{R^{3n}}$)

Cantor shows that such a bijection must always exist. While there are infinitely many such bijections, the easiest way would probably be to use Hilbert’s space filling curve. Each point along the curve corresponds to a unique circle in the desired output space, and it’s easy to define any point along the curve with a single scalar: how far along the curve is that point? If you continue the Hilbert curve until you’ve entirely tiled the output space, all possible circles will be accounted for.

It’s probably more practical to use 2 or 3 points to define the circle, as discussed in other answers. But you asked for the minimum :)

user577215664
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Matthew Bloomfield
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