Thread: Finding the plane of intersection of two spheres in Euclidian (3D) space.

so if two spheres intersect (one with centre (1,1,2) and radius R=4 and the other with centre (1,1,4) with radius r=2) (i just made these values up i hope they work). how can you find the PLANE that is defined by the intersection of both spheres (actually it's a circle but let's call it a plane)

The normal vector of the plane is obvious, it must be the vecto C1C2 (the vector between centre 1 and centre 2)

but how do you find the known point on the sphere then?


HELP very appreciated! thanks!

The point is on both spheres.

Originally Posted by Plato

The point is on both spheres.

no, not necessarily, that's only the case if bothe spheres are of equal radius; in this case they are not.

Maybe I am reading you example incorrectly.
It seems to me that the spheres are:
.
Did I read the question incorrectly? If not that point works.

Originally Posted by Yehia

so if two spheres intersect (one with centre (1,1,2) and radius R=4 and the other with centre (1,1,4) with radius r=2) (i just made these values up i hope they work). how can you find the PLANE that is defined by the intersection of both spheres (actually it's a circle but let's call it a plane)

The normal vector of the plane is obvious, it must be the vecto C1C2 (the vector between centre 1 and centre 2)

but how do you find the known point on the sphere then?


HELP very appreciated! thanks!

The small sphere (r = 2) lies completely inside of the large sphere (r = 4) touching the large sphere at T(1, 1, 6)

The first sphere has the equation:



The 2nd sphere has the equation:



Subtract both equations. You'll get:
which is the equation of the plane which contains the common points of both spheres. With your example the equation describes the tangent plane of both spheres.