# How many coordinates are necessary to determine a sphere?

Do determine a circle, you would need at least three coordinates. How many are necessary to determine a sphere?

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Hint: why you need three points to determine a circle. In other words, given three points, how do you determine the circle? – Patrick Li Nov 4 '12 at 17:16
I could take the bisections of two of the three points twice, and the intersection of those two lines would be the centre of the circle. The radius then is just the distance between the centre and one of the points. – Tim Vermeulen Nov 4 '12 at 17:20
So, to determine a sphere, I assume you would need three points again: taking the three bisections, see where those intersect and there's the centre of the circle. However, it doesn't seem logical to me a circle needs the same amount of coordinates to determine as a sphere. – Tim Vermeulen Nov 4 '12 at 17:24
In how many dimensions is your sphere embedded? – Peter Taylor Nov 4 '12 at 17:26
@timjver Note that the bisection of two points in three dimension is a plane, not a line. – Patrick Li Nov 4 '12 at 18:44

## 2 Answers

I believe the answer is 4 (and hopefully someone will correct me if I'm wrong...). Here's why. Any 3 points are coplanar. Consider 3 points on the plane A, B, and C. One can connect each pair of points with a line segment to create a triangle, then circumscribe a circle. The center of said circle is equidistant to all 3 points. Consider a line passing through the center perpendicular to this plane. It is easy to prove that any point on this line is also equidistant from A, B, and C. Therefore, it must take more than 3 points to determine a sphere.

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You're absolutely right. I at first thought three points are sufficient, because in order to find a point equidistant to the three given point, I would just have to see where the three planes bisecting the three points meet. (so, the bisection of A and B, of B and C, and of A and C.) However, these three planes don't meet in a single point, but in a line. That's why we need the fourth point. – Tim Vermeulen May 20 '13 at 18:19
@Mike: Not ANY line perpendicular to the plane.... You mean the line perpindicular to the plane and passing through the circumcenter of ABC. To see this, let the center of the sphere be O and the projection of O onto the plane containing ABC be O'. By the pythagorean theorem, the distance between O and A is sqrt((O'A)^2+(OO')^2) and similarly for B and C. Thus O'A=O'B=O'C, so O must lie on the perpindicular to the plane ABC passing through the circumcenter of ABC. From there it follows that at least 4 point on the surface of the sphere are needed to determine the center, as you said. – Joshua Benabou Jul 17 '15 at 2:45
@JoshuaBenabou The answer is corrected. I had meant for the line to pass through the center (my reason for mentioning it) and inadvertently left it out. Drawing in segments from A, B, and C to the center and any point P on the line results in 3 congruent right triangles: each share the leg from P to the center perpendicular to 3 radii of the circle. Therefore, the hypoteneuses, the line segments PA, PB, and PC, all have the same length. So any point on the line can be the center of a sphere containing A,B, and C. – Mike Jul 17 '15 at 3:45

A nice explanation for 4. 3 gives you a circle. But you need one more to determine the size of the circle compared to the sphere.

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