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Finding the centre and radius

  1. Mar 18, 2008 #1
    The question states:
    Find the centre and radius of each circle with equations as given
    (a) 3x^2 + 3y^2 = 81
    (b) x^2 = 6y - y^2

    I really don't know how to approach this question, i started (a) by dividing both sides by 3 but then i don't know where to go from there, and i don't even know how to do (b)...please help me:frown:
  2. jcsd
  3. Mar 18, 2008 #2


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    Staff Emeritus
    Science Advisor

    Well, you know that a circle with centre (p,q) and radius r can be written (x-p)2+(y-q)2=r2.

    (a) once you've divided through by 3, it is in the above form, isn't it?
    (b) try to put it in the above form by, maybe, completing the square for the y component.
  4. Mar 18, 2008 #3
    ok but cristo, for (a) you're saying that my centre would just be p and q?..no actual numbers?...and for (b) my radius would be 9? -also for (b) what would be my centre after completing the square for the y component?
  5. Mar 18, 2008 #4
    a) P and Q are dummy variables. In your first problem: 3x^2 + 3y^2 = 81 you could think of this as 3(x-0)^2 + 3(y-0)^2 = 81

    b) Remember the equation is (x-p)^2+(y-q)^2=r^2, make sure you're accounting for the radius^2.

    The center is straight forward once you have completed the square.
  6. Mar 18, 2008 #5
    ...thank u guys
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