Geometry of Sphere: Find All Points on Circle in Sphere

[kevinalm]

Subject wise, this seems the most appropriate forum, so I'll post here. Feel free to move.

Given a circle of radius r, whose center is distance R from point P and which always lies in a plane perpendicular to line defined by P and center of circle (r) at distance R. All points of the circle will always lie within the surface of a sphere centered on P of radius= sqr( r^2 +R^2). I'm pretty confident on my result but intuitively I didn't expect this.

 

[Andrew Mason]

Subject wise, this seems the most appropriate forum, so I'll post here. Feel free to move.

Given a circle of radius r, whose center is distance R from point P and which always lies in a plane perpendicular to line defined by P and center of circle (r) at distance R. All points of the circle will always lie within the surface of a sphere centered on P of radius= sqr( r^2 +R^2). I'm pretty confident on my result but intuitively I didn't expect this.

I am a little confused by your terms, but I think this is equivalent to slicing a sphere at a distance R from the centre. This generates a circular cross section and the points on the circumference of that circular cross section are all on the surface of the sphere.

AM

 

[kevinalm]

Yes, that is what I'm doing. I just came at it from a strange viewpoint. I was setting up some kinematic equations for a gyro rim, the axel of which is constrained to pass through point P, and the center of which is constrained to distance R from P. What surprised me was that for all r,R >0 all points of the gyro rim (idealized to a circle of course) lie within a sphere. On reflection, I see this is to be expected. Thanks.

 

[HallsofIvy]

Take any of the points, X, on the circle and draw the line from the center of the circle, O, to X, the line from O to P, and the line from X to P. Since the circle is in a plane perpendicular to OP, those three line segments form a right triangle. The distance from X to O is r and the distance from O to P is R. Use the Pythagorean to find the distance from X to P.

 

[kevinalm]

Yes, that was my original solution, in essence. I was looking at the "axel" rotating in the xy plane and was curious as to the nature of the surface of revolution. What I wasn't expecting was that for all non-negative R,r all points of circle (r) lie in a single sphere of radius= sqr(r^2 +R^2) . It even fails sensibly in the cases R=0, r=0 and R,r=0,0 collapsing to sphere radius r, sphere radius R and point P respectively. For some silly reason I was thinking of varying oblateness, depending on R,r. Didn't want to believe myself. Leads to some interesting implications I may explore, like the instantaneous v always in the tangent plane.(to the sphere at that point.)

Again Thanks.