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Thread: find the equations of tangent spheres of equal radii....

  1. #1
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    find the equations of tangent spheres of equal radii....

    Find the equations of the tangent spheres of equal radii, whose centers are (-3, 1, 2) and (5, -3, 6).

    I'm not quite sure if this is supposed to be two different equations or the same one since they have equal radii. It is also confusing me because it gives two centers instead of a diameter as was given in previous problems. Any help is greatly appreciated

    Thanks!

  2. #2
    The equation of sphere: (x-x0)^2+(y-y0)^2+(z-z0)^2=R^2
    where x0, y0, z0 - coordinates of center of the sphere
    So
    1-st sphere (x+3)^2+(y-1)^2+(z-2)^2=(d/2)^2
    2-nd sphere (x-5)^2+(y+3)^2+(z-6)^2=(d/2)^2
    where d is distance between two centers.

  3. #3
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    So even though it says they have equal radii it doesn't mean I set the two equations equal to each other and just have on equation for both spheres??

  4. #4
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    You'll have two equations, but can you find the radius?.

  5. #5
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    I don't think I can find the radius, so it would be fine for me to leave the R as the variable in the equation and keep the equations separtate?

    Thanks!

  6. #6
    Senior Member skeeter's Avatar
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    Quote Originally Posted by Erin0702
    I don't think I can find the radius, so it would be fine for me to leave the R as the variable in the equation and keep the equations separtate?

    Thanks!
    the radius of each sphere is just half the distance between the two centers ... can you find the distance between (-3, 1, 2) and (5, -3, 6)?

    [tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}[/tex]

  7. #7
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    Re: find the equations of tangent spheres of equal radii....

    Hello, Erin0702!

    You're making hard work out of a rather simple problem.
    There are no equations to solve . . .

    ]Find the equations of the tangent spheres of equal radii, whose centers are (-3, 1, 2) and (5, -3, 6).
    As guntram pointed out, the equation of the sphere with center [tex](x_o,\,y_o,\,z_o)[/tex] and radius [tex]r[/tex] is:

    [tex]\;\;\;(x\,-\,x_o)^2\,+\,(y\,-\,y_o)^2\,+\,(x\,-\,z_o)^2\;=\;r^2[/tex]


    You already know the centers . . . right?

    So we have: [tex]\,(x\,+\,3)^2\,+\,(y\,-\,1)^2\,+\,(x\,-\,2)^2\;=\;r^2[/tex]
    . . . . . . .and: [tex]\,(x\,-\,5)^2\,+\,(y\,+\,3)^2\,+\,(x\,-\,6)^2\;=\;r^2[/tex]

    and all we need is the radius.
    Code:
                 *             *
                    *       *
                      *   *
                       * *
    
         .- - - o - - - * - - - o
         (-3,1,2)   r       r   (5,-3,6)
                       * *
                      *   * 
                    *       *
                 *             *
    Since the radii are equal, the distance between the centers is twice the radius.

    Now go for it!
    I'm the other of the two guys who "do" homework.

  8. #8
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    ok thanks!

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