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Math Help - True or false?

  1. #1
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    True or false?

    x^2+y^2-2x-4y+6=0 is the equation of a circle.

    I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle x^2+y^2+2gx+2fy+c=0
    Thanks
    Last edited by arze; October 5th 2009 at 06:04 PM. Reason: latex
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  2. #2
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    Quote Originally Posted by arze View Post
    x^2+y^2-2x-4y+6=0 is the equation of a circle.

    I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle x^2+y^2+2gx+2fy+c=0
    Thanks
    x^2 - 2x + y^2 - 4y = -6

    complete the square for x and y ...

    x^2 - 2x + 1 + y^2 - 4y + 4 = -6 + 1 + 4

    (x-1)^2 + (y-2)^2 = -1

    since the general circle equation is

    (x-h)^2 + (y-k)^2 = r^2


    r^2 = -1 ?
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  3. #3
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    Quote Originally Posted by arze View Post
    x^2+y^2-2x-4y+6=0 is the equation of a circle.

    I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle x^2+y^2+2gx+2fy+c=0
    Thanks
    Actually, the "general equation" of a circle is of the form

    (x - h)^2 + (y - k)^2 = r^2

    which is a circle of radius r, centred at (h, k).


    I would try to transform the equation you have been given into this form.

    x^2 + y^2 - 2x - 4y + 6 = 0

    x^2 - 2x + y^2 - 4y = -6

    x^2 - 2x + (-1)^2 + y^2 - 4y + (-2)^2 = -6 + (-1)^2 + (-2)^2

    (x - 1)^2 + (y - 2)^2 = -1.


    Is r^2 = -1 possible?
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  4. #4
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    Ah!!! How could I have missed that. Thanks!
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