Babyslapmafro
  • Babyslapmafro
Please help me get started on the following problem. Find an equation of the sphere with center (2,-1,-3) satisfying the given condition: Tangent to the xy-plane.
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
If I recall correctly, a sphere with center \((h,k,l)\) has the equation \[r^2=(x-h)^2+(y-k)^2+(z-l)^2,\] where \(r\) is the radius of the sphere. You know the center, so you just have to find the radius of the sphere. It's tangent to the xy-plane, so you have to find the distance from the center (2,-1,-3) to this point's projection on the xy-plane, which would be (2, -1, 0).
Babyslapmafro
  • Babyslapmafro
How did you determine the point (2,-1,0)?
anonymous
  • anonymous
Forgive the poor drawing. It's a bit tough in three dimensions.|dw:1370989801438:dw|

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anonymous
  • anonymous
For every point in the xy-plane, there is no z-coordinate. In 3-space, you let \(z=0\), which (in the drawing) would move you "up" 3 units to the point (2, -1, 0): |dw:1370990076171:dw|
Babyslapmafro
  • Babyslapmafro
But we're dealing with a sphere
anonymous
  • anonymous
What's your point? A sphere is 3-dimensional, isn't it?
anonymous
  • anonymous
Perhaps you're confused by the lack of a sphere in the drawing? I only drew the sphere's center for simplicity, since that's all the information you really need. The center's distance from the xy-plane is what's needed to find the radius, and thus the equation of the sphere itself, as I said before.
Babyslapmafro
  • Babyslapmafro
ok so r^2=9 yes?
Babyslapmafro
  • Babyslapmafro
and is r positive or negative 3? Or is that irrelevant?
Babyslapmafro
  • Babyslapmafro
(x-2)^2+(y+1)^2+(z+3)^2=9

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