How do you write the equation of the circle in standard form? and what is the center and radius of it?

$\displaystyle 4x^2 + 4y^2 + 24x + 16y + 43 = 0$

A step-by-step view would really help to understand for future problems.

Thanks!

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- Nov 29th 2008, 07:58 PMmeiyukichanEquation of a Circle. HELP!
How do you write the equation of the circle in standard form? and what is the center and radius of it?

$\displaystyle 4x^2 + 4y^2 + 24x + 16y + 43 = 0$

A step-by-step view would really help to understand for future problems.

Thanks! - Nov 29th 2008, 10:40 PMmr fantastic
- Nov 29th 2008, 11:04 PMmeiyukichan
- Nov 29th 2008, 11:35 PMandreas
I will continue from here $\displaystyle 4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0$ (*)

$\displaystyle x^2+6x=x^2+6x+9-9=(x+3)^2 -9$**(1)**

$\displaystyle y^2+4y=y^2+4y+4-4=(y+2)^2-4$**(2)**

Substitute**(1)**and**(2)**into**(*)**:

you get $\displaystyle 4*(x+3)^2-4*9+4*(y+2)^2-4*4+43=4(x+3)^2+4(y+2)^2=9$ Center is at $\displaystyle x=-3$ and $\displaystyle y=-2$ ,Radius is**3**

- Nov 29th 2008, 11:36 PMmr fantastic
- Nov 30th 2008, 12:48 AMmeiyukichan
- Nov 30th 2008, 01:58 AMacevipa
$\displaystyle

4x^2 + 4y^2 + 24x + 16y + 43 = 0

$

Rearrange the equation:

$\displaystyle 4x^2 + 24x + 4y^2 + 16y = -43$

Try and make two perfect squares on the left side of the eqn.

$\displaystyle 4x^2 + 24x + 36 + 4y^2 + 16y + 16 = -43 + 36 + 16$

$\displaystyle (2x+6)^2 + (2y+4)^2 = 9$

Then find the centre and radius of the circle - Nov 30th 2008, 02:36 AMmr fantastic
- Nov 30th 2008, 09:43 AMmeiyukichan
- Nov 30th 2008, 11:06 AMmr fantastic
- Nov 30th 2008, 12:12 PMmeiyukichan
Oh! Thank you. I forgot that the equation was like this $\displaystyle (x - h)^2 + (y - k)^2 = r^2$

meaning that since it was a positive, it is a negative then...

I didn't read the insulting post, but if it was something like how can I make such a simply mistake and saying how stupid I am, then I don't care because I'm too young to understand this stuff fully anyways...

I'm only in the 8th grade...