# Rearranging an equation.

• Feb 19th 2010, 02:47 PM
MathBlaster47
Rearranging an equation.
A question I am working on asks me to find the center and radius of a circle with the equation: $x^2+y^2-10x+4y+13=0$.
The thing is, while I am pretty sure I know what to do to get my answers I'm having some trouble rearranging the equation to a workable format. I have one way that I think might work, but I've never really understood how that whole thing works.
Here goes!

$x^2+y^2-10x+4y+13=0 \rightarrow (x^2-10x)+(y^2+4y)=-13$

Am I on the right track?
• Feb 19th 2010, 02:52 PM
e^(i*pi)
Quote:

Originally Posted by MathBlaster47
A question I am working on asks me to find the center and radius of a circle with the equation: $x^2+y^2-10x+4y+13=0$.
The thing is, while I am pretty sure I know what to do to get my answers I'm having some trouble rearranging the equation to a workable format. I have one way that I think might work, but I've never really understood how that whole thing works.
Here goes!

$x^2+y^2-10x+4y+13=0 \rightarrow (x^2-10x)+(y^2+4y)=-13$

Am I on the right track?

You are, next complete the square for x and y

$(x+5)^2-25 + (y+2)^2 - 4 = -13$

Spoiler:
$(x+5)^2 + (y+2)^2 = 16$

Therefore we have a circle with centre (-5,-2) and radius 4
• Feb 19th 2010, 02:53 PM
icemanfan
You need to complete the square for both x and y.

Completing the square for

$x^2 - 10x$ yields $x^2 - 10x + 25 = (x - 5)^2$

and completing the square for

$y^2 + 4y$ yields $y^2 + 4y + 4 = (y + 2)^2$.

So we add $25 + 4 = 29$ to both sides of your equation:

$x^2 - 10x + 25 + y^2 + 4y + 4 + 13 = 29$

and then simplify:

$(x - 5)^2 + (y + 2)^2 = 16$

Can you find the answers from this form of the equation?
• Feb 19th 2010, 02:59 PM
MathBlaster47
Thanks!

Uhm...why are the numbers added always 25 and 4?
I had thought that they were determined by the equation at hand, I realize that isn't so, but my curiosity, as they say, has been piqued.
• Feb 19th 2010, 03:02 PM
icemanfan
Quote:

Originally Posted by MathBlaster47
Thanks!

Uhm...why are the numbers added always 25 and 4?
I had thought that they were determined by the equation at hand, I realize that isn't so, but my curiosity, as they say, has been piqued.

They are determined by the equation at hand.

To complete the square for $x^2 + ax$, you calculate

$\frac{a^2}{4}$ since

$x^2 + ax + \frac{a^2}{4} = \left(x + \frac{a}{2}\right)^2$

So these numbers are dependent on the coefficients of x and y.
• Feb 19th 2010, 03:03 PM
SuperCalculus
There's a much easier way.

You're working with a circle with equation with equation $x^2 + y^2 + 2gx + 2fy + c = 0$, where centre is $(-g, -f)$ and radius is $\sqrt{g^2 + f^2 - c}$

So, from the equation you gave, you can derive;
$-10x = 2gx$
and
$4y = 2fy$
Hence;

$g = -5$

$f = 2$

So, plug those values into the equations

Centre is $(5, -2)$

And radius is $\sqrt {25 + 4 - 13} = \sqrt {16} = 4$.
• Feb 19th 2010, 03:53 PM
MathBlaster47
My mistake...I had been looking at an example that also used the numbers 25 and 4, so I got a smidgen turned around, thank you for setting me straight!
• Feb 19th 2010, 06:42 PM
MathBlaster47
Quote:

Originally Posted by icemanfan
They are determined by the equation at hand.

To complete the square for $x^2 + ax$, you calculate

$\frac{a^2}{4}$ since

$x^2 + ax + \frac{a^2}{4} = \left(x + \frac{a}{2}\right)^2$

So these numbers are dependent on the coefficients of x and y.

Just so I'm clear, since there is no loose number all I have to do is the rest of the working for completing the square?

Quote:

Originally Posted by icemanfan
You need to complete the square for both x and y.

Completing the square for

$x^2 - 10x$ yields $x^2 - 10x + 25 = (x - 5)^2$

and completing the square for

$y^2 + 4y$ yields $y^2 + 4y + 4 = (y + 2)^2$.

So we add $25 + 4 = 29$ to both sides of your equation:

$x^2 - 10x + 25 + y^2 + 4y + 4 + 13 = 29$

and then simplify:

$(x - 5)^2 + (y + 2)^2 = 16$

Can you find the answers from this form of the equation?

Verily, thanks!

Quote:

Originally Posted by SuperCalculus
There's a much easier way.

You're working with a circle with equation with equation $x^2 + y^2 + 2gx + 2fy + c = 0$, where centre is $(-g, -f)$ and radius is $\sqrt{g^2 + f^2 - c}$

So, from the equation you gave, you can derive;
$-10x = 2gx$
and
$4y = 2fy$
Hence;

$g = -5$

$f = 2$

So, plug those values into the equations

Centre is $(5, -2)$

And radius is $\sqrt {25 + 4 - 13} = \sqrt {16} = 4$.

Thanks for the easy method! I'll remember it in the future!

Edit:
I, for some reason, am feeling a little lost as to how to simplify:
$(x^2-x10+25)+(y^2+4y+4)=16$
I just have to factor out the numbers right?
• Feb 19th 2010, 08:27 PM
Diagonal
OK, I can edit. How do I delete one of my own entries? Thanks.

I wasn't awake again, and should have looked through all the replied already done for him.
• Feb 20th 2010, 10:46 PM
mr fantastic
Quote:

Originally Posted by Diagonal
OK, I can edit. How do I delete one of my own entries? Thanks.

I wasn't awake again, and should have looked through all the replied already done for him.

You can't delete. But if you report a post to a Moderator using the Report Post Tool (the small triangle found in top right corner of all posts) you can request that a post be deleted. In fact, you can report any post that you feel deserves the attention of a Moderator.