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So if I have a circle in 3D space, of which I know just 4 points on the edge of the circle that are 90 degrees from each other. I want to know the maximum distance between this circle in 3D and a plane that lies directly below it on the same axis.

I'm not sure where to start I was thinking to try and find the equation of the circle but I don't know how to do that in 3D. Once we have that how do you find the max distance from any point in the circle to the plane below.

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  • $\begingroup$ In 3D, a circle can be expressed as the intersection between a sphere and a supporting plane. If you don't like intersections, I think it is possible to express it with a parametric equation, but it's a little annoying. Also, what do you mean by "a plane that lies directly below [the circle] on the same axis" ? The way I see it, your statement means that the plane that contains the circle, and your other plane, are parallel. The distance to the plane, would be constant for all points of the circle. So there's little interest in looking for a maximum/minimum. $\endgroup$ – N.Bach Nov 19 '17 at 16:02
  • $\begingroup$ On a side note, any 3 points that are not collinear suffice to uniquely define a circle. $\endgroup$ – N.Bach Nov 19 '17 at 16:04
  • $\begingroup$ Sorry I mean that the plane just lie below it is not necessarily parralllel with the circle $\endgroup$ – Sasha Nov 19 '17 at 16:05
  • $\begingroup$ Then it's probably equivalent to take any random plane. Your 4 points being on one side of the plane do not guarantee that the circle they lie on, and the plane, do not intersect. So if you just rotate the four points on your circle, it may very well happen that one of the points would suddenly be on the other of the plane, yet the situation seems basically identical to me. At least in terms of which points are on the circle, and which are on the plane. $\endgroup$ – N.Bach Nov 19 '17 at 16:12
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Let's call your $4$ points $X_1,X_2,X_3,X_4$, and the plane $\mathcal P$.

In your configuration, it is fairly easy to find the center and radius of your circle. Your four points indeed define two diameter of the circle. Let $C$ be the center.

Now that you know the center, you can easily find two orthogonal vectors that are parallel to the plane that contains the circle. Just pick an arbitrary point, say $X_1$, and some other point $X_i$ such that $X_1X_j$ is not a diameter. Then the two vectors $\vec{CX_1}$ and $\vec{CX_i}$ are orthogonal, with norm equal to the radius of the circle.

From there, a point $M$ is on the circle if and only if there exists a real number $\theta\in\mathbb R$ such that $$\vec{CM} = \cos(\theta) \vec{CX_1} + \sin(\theta) \vec{CX_i}$$

This gives you a parametric equation of the circle, from which you can probably derive the point on the circle with maximum distance to $\mathcal P$.


I personally don't really like the solution I suggested above, but I think it fits your initial attempt/idea better than what I had in mind initially.

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