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

• Mar 30th 2011, 10:40 AM
Yehia
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!
• Mar 30th 2011, 10:49 AM
Plato
The point \$\displaystyle (1,1,6)\$ is on both spheres.
• Mar 30th 2011, 10:54 AM
Yehia
Quote:

Originally Posted by Plato
The point \$\displaystyle (1,1,6)\$ 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.
• Mar 30th 2011, 11:01 AM
Plato
Maybe I am reading you example incorrectly.
It seems to me that the spheres are:\$\displaystyle (x-1)^2+(y-1)^2+(z-2)^2=16\$
\$\displaystyle \&~(x-1)^2+(y-1)^2+(z-4)^2=4\$.
Did I read the question incorrectly? If not that point works.
• Mar 30th 2011, 11:12 AM
earboth
Quote:

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:

\$\displaystyle (x-1)^2+(y-1)^2+(z-2)^2=16\$

The 2nd sphere has the equation:

\$\displaystyle (x-1)^2+(y-1)^2+(z-4)^2=4\$

Subtract both equations. You'll get: \$\displaystyle z - 6 = 0\$
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.