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Intersection of a sphere and plane

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Say I have a sphere with the equation:

    (x-4)^2 + (y-3)^2 + (z+2)^2 = 20

    How do I find the intersection with the xy-plane? Is the intersection at a point or it forms a circle

    2. Relevant equations



    3. The attempt at a solution

    Hmm.. I tried to set z equal to 0 to solve this problem, however I think that's not the way to solve this problem
     
  2. jcsd
  3. Jan 29, 2009 #2

    Dick

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    If you are in the xy plane, then z=0. What's wrong with setting z=0?
     
  4. Jan 29, 2009 #3
    then if I am asked to find the yz plane I set x to 0? the thing is then all intersection type will be a circle.. and I am not sure if that's the right answer
     
  5. Jan 29, 2009 #4

    Dick

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    The radius of the sphere is sqrt(20). The center of the sphere is (4,3,-2). The farthest coordinate plane is the yz plane. The center of the sphere is 4 units away from that. sqrt(20)>4. They are all circles.
     
  6. Jan 29, 2009 #5
    ok, so what if the question is

    "How does the sphere intersect each of the x-axis, y-axis, z-axis"?
     
  7. Jan 29, 2009 #6
    all of them? how do I find the equation though?
     
  8. Jan 29, 2009 #7

    Dick

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    I retracted that quick (and wrong) answer. For the x-axis you have to figure out if the circle in the x-y plane intersects the x-axis. More simply, where you hit the x-axis is where y=0, and z=0. Etc.
     
  9. Jan 29, 2009 #8
    if that's the case then what's left is just (x-4)^2 = 20, correct? in that case it will intersect at two points as it's a quadratic equation
     
  10. Jan 29, 2009 #9

    Dick

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    No, what's left if y=0 and z=0 is (x-4)^2+3^2+2^2=20.
     
  11. Jan 29, 2009 #10
    yes that's true, either way all will be in that form and solving it is just basically a quadratic equation and we get two points for each plane (xy-plane, yz-plane, xz-plane), correct?
     
  12. Jan 29, 2009 #11

    Dick

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    No. That's why I retracted my answer. The intersections with the z-axis have to satisfy (z+2)^2+3^2+4^2=20. Now what?
     
  13. Jan 29, 2009 #12
    z^2+4z+9 to make it more simple, right... and that is a quadratic equation, so why can't we get two points out of it?
     
  14. Jan 29, 2009 #13

    Dick

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    (z+2)^2=(-5) to put it even more simply. Some quadratic equation don't have any real solutions. Face it.
     
  15. Jan 29, 2009 #14
    oh, you're right.. we can't take the square root of 5... so for a special case for the z-axis, it doesn't intersect anywhere, not even at a point.. right?
     
  16. Jan 29, 2009 #15

    Dick

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    Right. You meant sqrt(-5), right?
     
  17. Jan 29, 2009 #16
    the equation with the intersection at the y-axis is simply:
    = (y-3)^2 = 0
    = y-3 = 0
    = y = 3

    so it just intersects at a point
     
    Last edited: Jan 30, 2009
  18. Jan 30, 2009 #17

    Dick

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    Sure.
     
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