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anonymous

  • 5 years ago

find an equation of the sphere with center (2,-6,4) and radius 5. Describe its intersection with each of the coordinates planes.

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  1. anonymous
    • 5 years ago
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    so i know how to do the equation, but how do i do the intersection?

  2. anonymous
    • 5 years ago
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    \[(x-2)^2 + (y+6)^2 + (z-4)^2 = 25\]

  3. Hero
    • 5 years ago
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    What course are you taking?

  4. anonymous
    • 5 years ago
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    calc 3

  5. anonymous
    • 5 years ago
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    the description of the intersection is an equation of a circle i believe based on the example prior to this problem

  6. amistre64
    • 5 years ago
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    |dw:1327252360580:dw|

  7. amistre64
    • 5 years ago
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    you know the distance you are from the plane; you know the Radius of the sphere, so we can calculate the radius of the circle in the plane

  8. amistre64
    • 5 years ago
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    the plane you pick is just dropping that variable from the picture to get your center in that plane: (1,2,3) centers in no-x plane at (2,3) the no-y plane at (1,3) and the no-z plane at (1,2)

  9. amistre64
    • 5 years ago
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    the distance away from the plane is the one that you dropped :)

  10. anonymous
    • 5 years ago
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    okay so the xy plane would be (x-2)^2 + *y+6)^2 = 9

  11. anonymous
    • 5 years ago
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    The sphere does not intersect the y=0 coordinate plane.

  12. anonymous
    • 5 years ago
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    wait it would be = 16?

  13. amistre64
    • 5 years ago
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    (2,-6,4) and radius 5 xy: no-z: center at (2,-6) with a distance of 4 radius in no-z = sqrt(5^2-4^2)

  14. anonymous
    • 5 years ago
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    ohhh okay okay

  15. anonymous
    • 5 years ago
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    wait still confused

  16. anonymous
    • 5 years ago
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    so what would be the equation for yz?

  17. Hero
    • 5 years ago
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    How do I unsubscribe from this?

  18. amistre64
    • 5 years ago
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    yz: no-x plane; centers on (y,z) with a radius of: sqrt(5^2 - x^2)

  19. anonymous
    • 5 years ago
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    so 25-4? root 21?

  20. amistre64
    • 5 years ago
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    yep

  21. amistre64
    • 5 years ago
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    and r^2 = 21 then in your equation of circle stuff

  22. anonymous
    • 5 years ago
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    but how would i do xz?

  23. anonymous
    • 5 years ago
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    r = root(25-36)

  24. amistre64
    • 5 years ago
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    well, it centers in the no-y so (x,z); its radius is then sqrt(5^2-y^2)

  25. amistre64
    • 5 years ago
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    yep

  26. anonymous
    • 5 years ago
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    but its a negative number

  27. amistre64
    • 5 years ago
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    it is isnt it :) tell me; if your stretching to reach 5 and your 6 away, do you hit it?

  28. anonymous
    • 5 years ago
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    yeah bc r =6?

  29. amistre64
    • 5 years ago
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    |dw:1327252967020:dw|

  30. anonymous
    • 5 years ago
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    oh so it never intersects?

  31. amistre64
    • 5 years ago
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    its negative meaning there is no real radius in that plane so we aint even touching it

  32. anonymous
    • 5 years ago
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    so there wouldn't be an equation for xz

  33. amistre64
    • 5 years ago
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    correct; the xz plane has no intersection with the sphere

  34. anonymous
    • 5 years ago
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    awesome! thank you so much!

  35. amistre64
    • 5 years ago
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    yep

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