# Equation of a Circle. HELP!

• Nov 29th 2008, 07:58 PM
meiyukichan
Equation 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?

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

A step-by-step view would really help to understand for future problems.
Thanks!
• Nov 29th 2008, 10:40 PM
mr fantastic
Quote:

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

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

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

Re-arange:
$4x^2 + 24x + 4y^2 + 16y + 43 = 0$

$\Rightarrow 4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0$.

Now complete the square in each bracket. You should see where to go from there.
• Nov 29th 2008, 11:04 PM
meiyukichan
Quote:

Originally Posted by mr fantastic
Re-arange:
$4x^2 + 24x + 4y^2 + 16y + 43 = 0$

$\Rightarrow 4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0$.

Now complete the square in each bracket. You should see where to go from there.

What do you mean by complete the square in each bracket? I don't understand...
• Nov 29th 2008, 11:35 PM
andreas
Quote:

Originally Posted by meiyukichan
What do you mean by complete the square in each bracket? I don't understand...

I will continue from here $4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0$ (*)

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

$y^2+4y=y^2+4y+4-4=(y+2)^2-4$ (2)

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

you get $4*(x+3)^2-4*9+4*(y+2)^2-4*4+43=4(x+3)^2+4(y+2)^2=9$ Center is at $x=-3$ and $y=-2$ ,Radius is 3
• Nov 29th 2008, 11:36 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Re-arange:
$4x^2 + 24x + 4y^2 + 16y + 43 = 0$

$\Rightarrow 4(x^2 + 6x) + 4(y^2 + 4y) + 43 = 0$.

Now complete the square in each bracket. You should see where to go from there.

Quote:

Originally Posted by meiyukichan
What do you mean by complete the square in each bracket? I don't understand...

The first bracket is $x^2 + 6x = (x + 3)^2 - 9$. Get it?
• Nov 30th 2008, 12:48 AM
meiyukichan
Quote:

Originally Posted by mr fantastic
The first bracket is $x^2 + 6x = (x + 3)^2 - 9$. Get it?

Then is this correct?

$4 (x - 3)^2 - 9 + 4 (y + 2)^2 - 4 + 43 = 0$

Rearrange: $4 (x - 3)^2 + 4 (y + 2)^2 - 4 + 43 -9 = 0$

$4 (x - 3)^2 + 4 (y + 2)^2 = -29$

Does that mean that the center would be (3,2)?
• Nov 30th 2008, 01:58 AM
acevipa
$
4x^2 + 4y^2 + 24x + 16y + 43 = 0
$

Rearrange the equation:

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

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

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

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

Then find the centre and radius of the circle
• Nov 30th 2008, 02:36 AM
mr fantastic
Quote:

Originally Posted by meiyukichan
Then is this correct?

$4 {\color{red}[} (x {\color{red}+} 3)^2 - 9{\color{red}]} + 4 {\color{red}[} (y + 2)^2 - 4 {\color{red}]}+ 43 = 0$

Rearrange: $4 (x {\color{red}+} 3)^2 + 4 (y + 2)^2 - {\color{red}16} + 43 - {\color{red}36} = 0$

$4 (x {\color{red}+} 3)^2 + 4 (y + 2)^2 = -29$ Mr F says: This should have told you that there were mistakes ..... How can the square of the radius be less than zero?

Does that mean that the center would be (3,2)?

Note the corrections I have made in red. The correction of the other mistake in the last line is left for you.
• Nov 30th 2008, 09:43 AM
meiyukichan
Quote:

Originally Posted by mr fantastic
Note the corrections I have made in red. The correction of the other mistake in the last line is left for you.

Then I should get:
$4 (x + 3^2) + 4(y + 2)^2 = 3^2$
...?

With the center being (3,2) and the radius 3...?

• Nov 30th 2008, 11:06 AM
mr fantastic
Quote:

Originally Posted by meiyukichan
Then I should get:
$4 (x + 3)^{\color{red}2} + 4(y + 2)^2 = 3^2$
...?

With the center being (3,2) and the radius 3...?

Apart from the typo, you're almost correct. The centre is at (-3, -2).

By the way, there was an insulting post which I have deleted. My apologies on behalf of MHF if you had already read it.
• Nov 30th 2008, 12:12 PM
meiyukichan
Quote:

Originally Posted by mr fantastic
Apart from the typo, you're almost correct. The centre is at (-3, -2).

By the way, there was an insulting post which I have deleted. My apologies on behalf of MHF if you had already read it.

Oh! Thank you. I forgot that the equation was like this $(x - h)^2 + (y - k)^2 = r^2$
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...