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    When you have to do completing the square to get the equation, can someone cofirm the negatives work like this

    so say if its x2 + y2 -10x + 6y - 15 = 0

    is it (x-5)2 (y+3) 2 - 9 -25 -15 = 0

    So you dont take into account the (x-a2) + (y-b2) - if i am making any sense?

    thank you.
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    (Original post by LondonBoy)
    When you have to do completing the square to get the equation, can someone cofirm the negatives work like this

    so say if its x2 + y2 -10x + 6y - 15 = 0

    is it (x-5)2 (y+3) 2 - 9 -25 -15 = 0

    That seems all right. So that would be (x-5)² + (y+3)² = 7²

    So you dont take into account the (x-a2) + (y-b2) - if i am making any sense?
    I think I know what you're talking about here. If the centre is, as it is here, say (3, -2) for example (with a negative co ordinate), you'd apply it to the formula as you'd do normally and then you'd get (x-3)² + (y+2)² ...
    I guess it just depends on the co ordinates of the centre and their signs.
 
 
 
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