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Thread: the circle equation

  1. #1
    Newbie
    Joined
    Jun 2007
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    9

    Unhappy the circle equation

    The circle with equation x^2 + 4x + y^2 - 6y = 3 has center C and radius r, where?

    • C = (0,0) ; r = 3
    • C = (2, -3) ; r = 16
    • C = (2, -3) ; r = 4
    • C = (-2,3) ; r = 16
    • C = (-2,3) ; r = 4
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  2. #2
    Junior Member
    Joined
    Jun 2007
    From
    Cambridge, UK
    Posts
    41
    The circle's equation is:

    $\displaystyle x^2 + 4x + y^2 - 6y = 3$

    Rearrange this using the "identities"
    $\displaystyle (x+2)^2 \equiv x^2 + 4x + 4$
    $\displaystyle (y-3)^2 \equiv y^2 - 6y + 9$

    and you'll find it changes to:

    $\displaystyle (x+2)^2 - 4 + (y-3)^2 - 9 = 3$
    $\displaystyle \Rightarrow (x+2)^2 + (y-3)^2 = 16$

    Compare it with the general form for a circle, radius $\displaystyle R$ centred at $\displaystyle (a,b)$:

    $\displaystyle (x-a)^2 + (x-b)^2 = R^2$

    and you can see that $\displaystyle R =4$, and the centre is $\displaystyle (-2, 3)$.
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