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Suppose, two spheres intersect. Subtracting the equations of the speheres, a linear equation appears which indicates the plane conataining all points belonging to the intersection of the spheres.

But the intersection is only a small part of this plane.

  • Can this intersection be described ? Do the points form a well-known geometric shape ?
Peter
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  • Yes, I edited the qeustion. – Peter May 10 '15 at 23:38
  • just intersect the plane with either of the spheres – WW1 May 10 '15 at 23:38
  • I do not want to calculate the intersecting points. I want to know if the intersection can be visualized. – Peter May 10 '15 at 23:40
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    Working in Cartestian coordinates, suppose you have 2 spheres lying in $\mathbb{R}^3$: $(x-a_i)^2 + (y-b_i)^2 + (z-c_i)^2 = r_i^2,$ where $i=1,2.$ Expanding the square terms, and subtracting you see the intersection lies in a plane. However, the actual intersection is a circle that happens to lie in the plane generated above. See http://mathworld.wolfram.com/Sphere-SphereIntersection.html – matt biesecker May 10 '15 at 23:42
  • Intuitively, if you're convinced that their intersection lies on a plane, is it a much greater leap to intuitively believe that the intersection must be circular, as that's the only way a plane can intersect a sphere? – pjs36 May 11 '15 at 00:51
  • The intersection of $F(x,y,z)=0$ and $G(x,y,z)=0$ can be written as $(F(x,y,z))^2+(G(x,y,z))^2=0$. The union can be written as $(F(x,y,z))(G(x,y,z))=0$. – Akiva Weinberger May 11 '15 at 01:26

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You can always choose a coordinate system such that the first sphere is the larger one (if radii differ) and is centered around the origin: $$ x^2 + y^2 + z^2 = R^2 \quad (*) $$ and that the second sphere is centered around $(0,0, h)$: $$ x^2 + y^2 + (z - h)^2 = r^2 \quad (**) $$ for an intersection we should have $h \le R + r$.

For the intersection both equations $(*)$ and $(**)$ must hold. Substracting $(*)$ from $(**)$ we get:

$$ -2zh+h^2 = r^2-R^2 \\ z = \frac{h^2 + R^2 - r^2}{2h} =: z_i $$

The equation $(*)$ then reduces to $$ x^2 + y^2 = R^2 - z_i^2 = r_i^2 \quad (\#) $$ with $$ r_i = \sqrt{R^2 - z_i^2} $$ The equation $(\#)$ of the intersection is the equation of a circle with radius $r_i$.

mvw
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