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Spheres and planes

  1. Dec 14, 2012 #1
    Please bear with my ignorance. I will try to explain the complete scenario . I have a 3D cuboid (with planes as front, back, left, right, top and bottom) and three spheres called s1, s2 and s3. s1's center is at (2,4,5) and radius is 2. s2's center is at (-2,3,2) and radius is 1. s3's center is at (2,4,-2) and radius is 3.

    Now I want to find where these spheres will intersect the planes and also find the center and radii of circles they will make on the planes. The coordinates of the cuboid walls are (bottom left, bottom right, top right, top left) as follows:

    (-2, 0, -4),(2, 0, -4),(2, 3, -4),(-2, 3, -4)

    (-2, 0, 4),(-2, 0, -4),(-2, 3, -4),(-2, 3, 4)

    (2, 0, -4),(2, 0, 4),(2, 3, 4),(2, 3, -4)

    (2, 0, 4),(-2, 0, 4),(-2, 3, 4),(2, 3, 4)

    (-2, 3, -4),(2, 3, -4),(2, 3, 4),(-2, 3, 4)

    (-2, 0, 4),(2, 0, 4),(2, 0, -4),(-2, 0, -4)

    I hope the question is clear enough to understand. Thank you.
  2. jcsd
  3. Dec 14, 2012 #2


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    This plane has equation z= -4. The first sphere has equation [itex](x-2)^2+ (y-4)^2+ (z- 5)^2= 4[/itex]. Setting z= 4, [itex](x- 2)^2+ (y- 4)^2+ 81= 4[/itex]

  4. Dec 14, 2012 #3
    Thanks for the help. Now how should I find the projected circle from the below equation:

    (x−2)^ 2 +(y−4)^ 2 +81=4

    I need the circles center and also its radius on the plane. Thanks.
  5. Dec 14, 2012 #4


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    There are no real solutions to that equation.

    What do you think that means?
  6. Dec 15, 2012 #5
    I know the planes' coordinates, so I should be able to find the projection of the sphere on these planes (the circles that the sphere creates on the planes on intersection).
  7. Dec 15, 2012 #6


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    "The circles that the sphere creates on the planes on intersection" are NOT the same as "the projection of the sphere on these planes". They are, as I said before, given by putting the equation of the plane into the equation of the sphere and finding which point are on the plane and the sphere- that is, the equation of the circle of intereserction. And SammyS has told you that the first equation, using the first plane, does NOT have any real solution. What does that tell you about the circle of intersection?

    Do you not know the general equation of a circle? It is [itex](x- a)^2+ (y- b)^2= R^2[/itex] where (a, b) is the center and R is the radius.
  8. Dec 15, 2012 #7
    Thank you all for all the help so far. To be honest I am now more confused than I was before, as you are now saying that the projections of the spheres are not the same as the circles created by these spheres on the planes. If you could be kind enough to point me to some online tutorial or material that could explain this to me in detail. I would really appreciate that as I dont want to ask any further questions being totally lost (the questions would only annoy you all and make me look more foolish :-) Thank you.
  9. Dec 15, 2012 #8


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    We are here to help. What tends to annoy us, and also makes those who as the questions look foolish, is when the questioner ignores the questions and advice of those who are trying to help -- merely gives responses without much thought. You don't appear to fit that description.

    An example of the difference between, 1) the projection of a sphere onto a plane, and 2) the intersection of a sphere with a plane :

    Consider your 3rd sphere, ( radius of 3 units, centered at (2,4,-2) ) and the coordinate planes. BTW: This sphere has the equation, (x-2)2 + (y-4)2 + (z+2)2 = 9

    This sphere does not intersect the xz-plane at all, because its center is at y=4, and its radius is 3. However, the projection of this sphere onto the xz-plane is a circle with a radius of 3 units, and a center at (2,0,-2). The equation of this circle is (x-2)2 + (z+2)2 = 9, y=0 .

    This sphere does intersect the other two coordinate planes. Its intersection with the xy-plane is given by (x-2)2 + (y-4)2 + (0+2)2 = 9 , which is equivalent to (x-2)2 + (y-4)2 = 5 and of course z = 0 . This circle has radius, √5, and is centered at (2,4,0) and lies in the xy-plane. The projection of the sphere onto the xy-plane is also a circle. This circle has the same center but its radius is 3.

    BTW: The projection of the sphere onto any plane is a circle of radius 3.

    I hope this helps.
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