How do I solve centre of sphere

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We know that coordinates of two points on sphere and radius ,how to get coordinate of centre ?
thank you!

Posted 30-Dec-14 15:28pm

Member 11344163

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PIEBALDconsult 30-Dec-14 21:35pm

As far as I know, two isn't enough, but I don't think there is a set minimum either -- in a worst-case you could get a bunch of points that form a circle but not know anything other than a minimum size of the sphere.

Sergey Alexandrovich Kryukov 30-Dec-14 22:41pm

Totally wrong. You don't get minimal size of the sphere, because the radius is known, which defines the size exactly. But the problem has infinite number of solutions in general position, they all lie on some circle which is easy to find; the special case when the solution is only one is when the distance between those points is exactly the equal to double radius.
I depicted a solution, please see Solution 1.
—SA

PIEBALDconsult 30-Dec-14 22:58pm

Oh, I see he says radius, but there are still two possible solutions.

Sergey Alexandrovich Kryukov 30-Dec-14 23:17pm

Not two. In 3D, its zero, exactly one, or continuum of solutions. Everything is explained in my answer. Is it unclear?
—SA

PIEBALDconsult 30-Dec-14 23:19pm

"Is it unclear?"

To me, yes, but maybe not to others.

Sergey Alexandrovich Kryukov 31-Dec-14 0:25am

Well, it's hard to explain it perfectly clear and graphical, with so limited graphic capabilities. I wish we could use HTML5-embedded SVG, but this site does not allow rendering it.
—SA

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Sergey Alexandrovich Kryukov · Answer 1 · 2014-12-30T16:50:00

Solution 1

Please see my comment to the comment to the question by PIEBALDconsult. Picture this:

 p1      p2
 \      /
r \    / r
   \  /
    \/
     center

This is how the location of the center looks in general case, when you have the continuum of solutions. Apparently, the center of the sphere of radius r lies on the sphere surrounding each of the points, p1 as a center, or p2 as a center. The two spheres intersect forming a circle (perpendicular to the plane of my drawing, with the center which lies symmetrically between p1 and p2). Each point of this circle is a solution.

If the distance between p1 and p2 is exactly 2*r, you have only one solution:

p1                       p2
------------+------------
          center 
     r              r

Finally, if the distance between p1 and p2 is greater that 2*r, there are no solutions.

This is an elementary problem on high-school geometry, in fact, off-topic.

—SA

Posted 30-Dec-14 16:50pm

Sergey Alexandrovich Kryukov

Comments

PIEBALDconsult 30-Dec-14 22:58pm

You're assuming they're on a great circle of the sphere? I'm not.
(OK, two points would be I suppose, but three might not be. I'll just shut up now.)

Sergey Alexandrovich Kryukov 30-Dec-14 23:18pm

...

PIEBALDconsult 30-Dec-14 23:24pm

The question is perfectly clear; the only problem is having zero or multiple solutions, but, in wider sense of the word "solution", this is a solution. Of course, it's very likely that the "problem" was formulated without understanding of all this, anyway, this is not a problem at all. In this sense, yes, it is not good.
—SA

Sergey Alexandrovich Kryukov 30-Dec-14 23:42pm

The question is perfectly clear; the only problem is having zero or multiple solutions, but, in wider sense of the word "solution", this is a solution. Of course, it's very likely that the "problem" was formulated without understanding of all this, anyway, this is not a problem at all...
—SA

Sergey Alexandrovich Kryukov 30-Dec-14 23:42pm

The question is perfectly clear; the only problem is having zero or multiple solutions, but, in wider sense of the word "solution", this is a solution. Of course, it's very likely that the "problem" was formulated without understanding of all this, anyway, this is not a problem at all. In this sense, yes, it is not good.
—SA

Sergey Alexandrovich Kryukov 30-Dec-14 23:19pm

No, I'm not assuming anything. Please read carefully. What are we talking about? This is the exercise for kids...
—SA

[no name] 31-Dec-14 2:22am

Indeed. The centres of all spheres with radius greater than half the distance between the two points and whose surface intersects the two points lie on a plane perpendicular to and half way along the line joining the two points. The OP has referred to a the case where the radius is known and the locus of points forming the solution is a circle on this plane with it's centre on the line joining the two points.

Snesh Prajapati · Answer 2 · 2014-12-30T17:06:00

Solution 2

For a sphere, just by knowing two coordinates of its surface and its radius - getting coordinate of its center is not possible.

Even for circle to get the coordinate of its center - we need at least three points on its circumference. and if you just know two points of circumference and radius of circle - there would be two possible circles with that data. Please have a look on below link:
http://mathforum.org/library/drmath/view/54490.html[^]

To figure out center of a sphere, you need at least four points, more specifically at least four noncoplanar points, and contains three noncollinear points . There is discussion about the same here:
http://steve.hollasch.net/cgindex/geometry/sphere4pts.html[^]

Hope this will help. Thanks.

Posted 30-Dec-14 17:06pm

Snesh Prajapati

Updated 30-Dec-14 18:41pm

v6

Comments

Sergey Alexandrovich Kryukov 30-Dec-14 23:21pm

You completely missed the fact that known are not two points, but two points and radius. The problem is well too simple, I provided complete analysis in Solution 1.
—SA

Snesh Prajapati 31-Dec-14 0:28am

I am writing in my first sentence of above answer "knowing two coordinates of its surface and its radius"...I do not understand what you are trying to say here.

Sergey Alexandrovich Kryukov 31-Dec-14 0:54am

I say that you are trying to apply considerations which are relevant to a different problem, when the points on the circle or sphere are known. For the problem formulated by OP, they are inapplicable. Mentioning the radius does not mean that your conclusions are correct.

The problem is too simple to discuss much. It has either zero, exactly one or continuum of solutions, depending on the distance between points compared to the radius. I explained it all in my answer.

—SA

PIEBALDconsult 30-Dec-14 23:23pm

"you need at least four points"
What if they're all on the same non-great circle? Wouldn't there still be two possible solutions?

Snesh Prajapati 31-Dec-14 0:40am

You are right..I update my answer by adding more specific words. Thanks.