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# Circle Equation watch

1. 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.
2. (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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