# Thread: Find the coordinates of the center and the radius of the circle

1. ## Find the coordinates of the center and the radius of the circle

Hey, i'm trying to solve this and im confused by the 2x and 2y. How do I treat this problem?

Thanks guys.

2. Originally Posted by bigitme
Hey, i'm trying to solve this and im confused by the 2x and 2y. How do I treat this problem?

Thanks guys.
$\displaystyle 2x^2+2y^2+18x+4=0$

First divide the equation through by 2:

$\displaystyle x^2+y^2+9x+2=0$

Subtract 2 from both sides:

$\displaystyle x^2+y^2+9x=-2$

Group the x terms together:

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

Complete the square:

$\displaystyle \left(x^2+9x+\tfrac{81}{4}\right)+y^2=-2+\tfrac{81}{4}$

$\displaystyle \left(x+\tfrac{9}{2}\right)^2+y^2=\tfrac{77}{4}$

Thus, the center of the circle is ........

The radius of the circle is ........

I hope this helps!

--Chris

3. Originally Posted by bigitme
Hey, i'm trying to solve this and im confused by the 2x and 2y. How do I treat this problem?

Thanks guys.

$\displaystyle 2x^2+2y^2+18x+4=0$

divide by $\displaystyle 2$:

$\displaystyle x^2+y^2+9x+2=0$

Complete the square for $\displaystyle x$:

$\displaystyle (x+4.5)^2+y^2+2-4.5^2=0$

so you have:

$\displaystyle (x+4.5)^2+y^2=4.5^2-2$

RonL

4. Originally Posted by Chris L T521
$\displaystyle 2x^2+2y^2+18x+4=0$

First divide the equation through by 2:

$\displaystyle x^2+y^2+9x+2=0$

Subtract 2 from both sides:

$\displaystyle x^2+y^2+9x=-2$

Group the x terms together:

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

Complete the square:

$\displaystyle \left(x^2+9x+\tfrac{81}{4}\right)+y^2=-2+\tfrac{81}{4}$

$\displaystyle \left(x+\tfrac{9}{2}\right)^2+y^2=\tfrac{7{\color{ red}3}}{4}$

Thus, the center of the circle is ........

The radius of the circle is ........

I hope this helps!

--Chris
Typo!

5. Originally Posted by masters
Typo!
argh! Thanks for catching that

--Chris