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Coordinate geometry help - C2

I know you have to get it into the formate (x-a)^2+(y-b)^2=r^2

Then you can work out the coordinates and use Pythagorus theorem to find out the radius

any help ?

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You seem to know the method, what have you done so far?

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Original post by User32432432

I know you have to get it into the formate (x-a)^2+(y-b)^2=r^2

Then you can work out the coordinates and use Pythagorus theorem to find out the radius

any help ?

Do you know how to complete the square?

OP

Original post by raiden95

You seem to know the method, what have you done so far?

dont really know were to start

OP

Original post by Mr M

Do you know how to complete the square?

yes sir

Original post by User32432432

dont really know were to start

Try completing the square using all terms with y as a start

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Original post by User32432432

yes sir

Can you complete the square on this bit?

y^2-6y

Original post by Fool In The Rain

If you put it in the form you mentioned, you should get:
x^2+(y-3)^2-7=0
That will give you your coordinates.

Then you expand the brackets, that will help you find the radius. (I'm not 100% sure though)

No this is wrong, give the OP a chance to try it

Original post by User32432432

I know you have to get it into the formate (x-a)^2+(y-b)^2=r^2

Then you can work out the coordinates and use Pythagorus theorem to find out the radius

any help ?

complete square

arrange into circle equation format

take the square root of r^2

write down answer

Job done

OP

Original post by Mr M

Can you complete the square on this bit?

y^2-6y

(y-3y)^2 +9

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Original post by User32432432

(y-3y)^2 +9

No that's very wrong.

Before you can tackle this question you need to go away and revise completing the square.

x^2 + bx = 0. [br][br](x+b/2)^2-(b/2)^2 = 0

Is this any help?

(edited 11 years ago)

OP

Original post by raiden95

x^2 + bx = 0. [br][br](x+b/2)^2-(b/2)^2 = 0

Is this any help?

thats what I did ?

(x+b/2)^2-(b/2)^2 = 0

You didn't do that.

(x+bx/2)^2+(b/2)^2 = 0

You did this.

(edited 11 years ago)

Study Forum Helper

Original post by raiden95

(x+b/2)^2-(b/2)^2 = 0

You didn't do that.

(x+b/2)^2+(b/2)^2 = 0

You did this.

No he did this:

(x+\frac{b}{2} x)^2+(\frac{b}{2})^2 = 0

Original post by Mr M

No he did this:

(x+\frac{b}{2} x)^2+(\frac{b}{2})^2 = 0

Only just noticed :colondollar:

:colondollar:

just complete the square

OP

Original post by Mr M

No he did this:

(x+\frac{b}{2} x)^2+(\frac{b}{2})^2 = 0

Sorry for the late reply, is this correct ?
(y+3)^2+9y

Original post by User32432432

Sorry for the late reply, is this correct ?
(y+3)^2+9y

This is wrong, you need to look at completing the square in your maths text book

OP

Original post by raiden95

This is wrong, you need to look at completing the square in your maths text book

Can you just explain what I have done wrong ?

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