Find the coordinates of the center and the radius of the circle

**bigitme** · September 11th 2008, 09:03 AM

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

Thanks guys.

**Chris L T521** · September 11th 2008, 09:09 AM

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.

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

First divide the equation through by 2:

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

Subtract 2 from both sides:

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

Group the x terms together:

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

Complete the square:

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

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

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

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

I hope this helps!

--Chris

**CaptainBlack** · September 11th 2008, 09:13 AM

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.

Your equation is:

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

divide by $2$ :

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

Complete the square for $x$ :

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

so you have:

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

RonL

**masters** · September 11th 2008, 09:15 AM

Originally Posted by Chris L T521

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

First divide the equation through by 2:

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

Subtract 2 from both sides:

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

Group the x terms together:

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

Complete the square:

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

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

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

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

I hope this helps!

--Chris

Typo!

**Chris L T521** · September 11th 2008, 09:20 AM

Originally Posted by masters

Typo!

argh! Thanks for catching that

--Chris

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