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Coordinate geometry help - C2

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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.
Original post by User32432432
Can you just explain what I have done wrong ?


y2−6y=(y−3)2−9y^2 - 6y = (y-3)^2 -9
Reply 22
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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 '-'
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.
Reply 24
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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
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 r2r^2
Original post by User32432432
What do I do after this stage ? (y-3)^2 - 9


Replace y2−6yy^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=r2(x-a)^2 +(y-b)^2=r^2 has centre (a, b) and radius r
Original post by Indeterminate
Do the same with the x terms.


There is nothing to do.
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).
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 :tongue:
Reply 30
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Original post by Mr M
Replace y2−6yy^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=r2(x-a)^2 +(y-b)^2=r^2 has centre (a, b) and radius r


Thanks!
is r^2=16?
Reply 31
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Original post by User32432432
Thanks!
is r^2=16?


Yes, well done :biggrin:
Reply 32
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Original post by raiden95
Yes, well done :biggrin:

Finally lol

Thanks for your help, really appreciate it

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