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    (Original post by User32432432)
    Sorry for the late reply, is this correct ?
    (y+3)^2+9y
    No you really must follow my advice and revise completing the square. Guessing like this is not helping your understanding.
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    (Original post by User32432432)
    Can you just explain what I have done wrong ?
    y^2 - 6y = (y-3)^2 -9
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    (Original post by Mr M)
    No you really must follow my advice and revise completing the square. Guessing like this is not helping your understanding.
    Oh, I understand now, its just in my notes I was working it out as if it was y^2+6y, but I only just realised its '-'
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    (Original post by User32432432)
    Oh, I understand now, its just in my notes I was working it out as if it was y^2+6y, but I only just realised its '-'
    That isn't your only problem with this. You are trying to add another term instead of subtract it and you have made this term a multiple of y when it isn't.
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    (Original post by Mr M)
    That isn't your only problem with this. You are trying to add another term instead of subtract it and you have made this term a multiple of y when it isn't.
    What do I do after this stage ? (y-3)^2 - 9
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    (Original post by User32432432)
    What do I do after this stage ? (y-3)^2 - 9
    Do the same with the x terms.

    Note that the constant terms (i.e numbers) form the radius squared, namely r^2
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    (Original post by User32432432)
    What do I do after this stage ? (y-3)^2 - 9
    Replace y^2 - 6y in the original equation and you have the equation of the circle in completed square form. This allows you to state the coordinates of the centre and the radius.

    (x-a)^2 +(y-b)^2=r^2 has centre (a, b) and radius r
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    (Original post by Indeterminate)
    Do the same with the x terms.
    There is nothing to do.
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    I think I know where I went wrong before when attempting to answer this (deleted it because it was wrong, I don't want anybody looking at it thinking it was right).
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    (Original post by Mr M)
    There is nothing to do.
    A bit foolish on my part to assume that he'd not dealt with the x terms already
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    (Original post by Mr M)
    Replace y^2 - 6y in the original equation and you have the equation of the circle in completed square form. This allows you to state the coordinates of the centre and the radius.

    (x-a)^2 +(y-b)^2=r^2 has centre (a, b) and radius r
    Thanks!
    is r^2=16?
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    (Original post by User32432432)
    Thanks!
    is r^2=16?
    Yes, well done
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    (Original post by raiden95)
    Yes, well done
    Finally lol

    Thanks for your help, really appreciate it
 
 
 
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