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Set of points specified by x^2 + y^2 <= 4x + 4y

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data

    What set of points is specified by the inequality x^2 + y^2 ≤ 4x + 4y

    2. Relevant equations

    x^2 + y^2 = r^2 is the formula for a circle with its center at the origin

    3. The attempt at a solution

    x^2 - 4x + y^2 - 4y ≤ 0 ??????


    The book gives the solution, if you want me to post it i can. But i didn't understand how they got it
     
    Last edited: Dec 16, 2011
  2. jcsd
  3. Dec 16, 2011 #2

    Mark44

    Staff: Mentor

    This is actually an inequality, not an equation.
    Complete the square in the x terms and in the y terms. The < part of the inequality represents all of the points inside a circle. The = part represents all the points on the circle.
     
  4. Dec 16, 2011 #3
    ninja edited :wink:

    are we completing the square to get the inequality in the familiar (x-a)^2 + (y-a)^2 = R^2 form that represents a circle? So it would be (x-2)^2 + (y-2)^2 ≤ 8

    So apparently that means the same thing as (x-0)^2 + (y-0)^2 ≤ 4x + 4y, even though (x-0)^2 + (y-0)^2 would indicate that the circle is at the origin whereas (x-2)^2 + (y-2)^2 is for a circle with the center at (2,2).

    Could someone help me understand why setting (x-0)^2 + (y-0)^2 less than or equal to "4x + 4y" rather than a number can make the circle centered at (2,2) instead of the origin as (x-0) and (y-0) led me to believe?
     
  5. Dec 16, 2011 #4

    Mark44

    Staff: Mentor

    Yes. This inequality can be separated into two statements:
    (x-2)^2 + (y-2)^2 < 8
    (x-2)^2 + (y-2)^2 = 8
    The inequality represents all the point inside the circle.
    The equation represent all the points on the circle.

    Together, the ≤ represents all the points on the circle or inside it.
    Yes, but this form is not helpful at all.
     
  6. Dec 16, 2011 #5

    Thanks Mark44!
     
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