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Question relating to shifting a circle.

  1. Jul 2, 2012 #1
    This problem arose for me while working out a triple integral in spherical coordinates. Basicly I know that when we shift a parabola along the axis it is simply translated. I naturally assumed that if we shifted a circle in a similar manner that it would act the same.

    However when we shift a circle along the axis, such as one with the equation

    [itex] (x-1)^2 + y^2 = 1 [/itex] We find that the entirety of the circle now sits above the x axis, and that our radius becomes [itex] 2cos∂ [/itex]

    Could anyone shed some light on this?
     
  2. jcsd
  3. Jul 2, 2012 #2

    jedishrfu

    Staff: Mentor

    you've moved the center to (1,0) but the circle should still have a radius of 1

    try some points (0,0) , (1,1) , (1,1) and (1,-1) all satisfy the equation and show that the radius is 1 and not as you say and that ithe circle still lies on the x-axis
     
  4. Jul 2, 2012 #3

    SammyS

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    [itex] (x-1)^2 + y^2 = 1 [/itex] is the equation of a circle having radius 1 with its center at (1, 0).

    Here's a ploy from WolframAlpha:
    attachment.php?attachmentid=48823&stc=1&d=1341241390.gif
     

    Attached Files:

  5. Jul 2, 2012 #4
    Alright I guess I just misunderstood the solution on the problem set.. thank you.
     
  6. Jul 2, 2012 #5

    SammyS

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    After reading your Original Post, it looks as if you might be converting [itex](x-1)^2 + y^2 = 1[/itex] to polar coordinates with the result, [itex] r=2\cos\theta\,. [/itex] If so, that variable, r, does not refer to the radius of the circle, it's the the distance that the point (x, y) is from the origin.
     
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