Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Center of a Sphere in R^3

  1. Sep 10, 2007 #1
    Center of a Sphere in R^3

    I need to find the center and radius of a sphere given the equation:

    [tex]
    x^2 - 4x + y^2 + z^2 = 0
    [/tex]

    I would like to hope it would be as easy as just adding 4x to both sides, but having a variable as the radius probably isn’t correct.

    On the other hand, I have no idea how to complete the square when no other coefficients are present. Simply adding y and z coefficients only leads to variables I cannot get rid of on the right side of the equation.

    Attempts I have thus far
    [tex]
    (x-2)^2 + (y+2)^2 + (z+2)^2 = 12 + 4y + 4z
    [/tex]
    or
    [tex]
    (x-2)^2+y^2+z^2=(\frac{4}{2})^2
    [/tex]
    I’m guessing I need to somehow get x^2 on the left and have the sphere centered about the origin.

    Any help would be great, thanks.
     
  2. jcsd
  3. Sep 10, 2007 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You'll have to complete the square for the x variables then the answer should be straight forward. The cartesian equation of a square centered at (x0, y0, z0) is:

    [tex] (x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2= r^2 [/tex]
     
  4. Sep 10, 2007 #3

    Avodyne

    User Avatar
    Science Advisor

    If I give you the equation (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2, where a, b, c, and r are constants, does it describe a sphere? (Answer: yes.) What are the (x,y,z) coordinates of the center of the sphere? What is the radius?
     
  5. Sep 10, 2007 #4
    So would it then be
    [tex]
    (x-2)^2+(y-0)^2+(z-0)^2=(\frac{4}{2})^2
    [/tex]?

    That's about all I can think of =(
     
  6. Sep 10, 2007 #5

    Avodyne

    User Avatar
    Science Advisor

    OK, so what is the center? What is the radius?
     
  7. Sep 11, 2007 #6
    If that's the equation for the sphere, the center would be (2,0,0) with r=2
     
  8. Sep 11, 2007 #7

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Correct! :smile: Although I don't understand where you got 4/2 from?
     
  9. Sep 11, 2007 #8
    The 4/2 was the factor that I added when I completed the square on both sides. I just left it that way because I wasnt sure where to go from there.

    Thank you both for your help!
     
  10. Sep 11, 2007 #9

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I only got a factor of 4. Never mind you have the correct answer anyway.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook