# Thread: Complete the square to find center and radius of the sphere

1. ## Complete the square to find center and radius of the sphere

I have an odd problem in my book that i am trying to solve but dont have the solution to.
Can anyone help me?

Find the center and radius of the sphere whose equation is:

$\displaystyle x^2 + 2x +y^2 - y + z^2 = 0$

Attempt:

$\displaystyle (x^2 + 2x + 1) -1 (y^2 - y - 1/4) + 1/4 + z^2 = 0$
$\displaystyle (x^2+1)^2 + ( y^2 - 1/2)^2 + z^2 = 1 - 1/4 = 3/4$

r = $\displaystyle r^2 = 3/4$
r=$\displaystyle \sqrt {3/4}$

radius: $\displaystyle 2 \sqrt {3}$

Center: $\displaystyle (-1, + 1/2, 0)$

Am i correct?

2. ## Re: Complete the square to find center and radius of the sphere

You are not correct with:

$\displaystyle r=\sqrt{\frac{3}{4}}$

but even if you were, this simplifies to:

$\displaystyle r=\frac{\sqrt{3}}{2}$

3. ## Re: Complete the square to find center and radius of the sphere

Originally Posted by icelated
I have an odd problem in my book that i am trying to solve but dont have the solution to.
Can anyone help me?

Find the center and radius of the sphere whose equation is:

$\displaystyle x^2 + 2x +y^2 - y + z^2 = 0$

Attempt:

$\displaystyle (x^2 + 2x + 1) -1 (y^2 - y - 1/4) + 1/4 + z^2 = 0$
Yes, $\displaystyle x^2+ 2x+1= (x+ 1)^2$ is a perfect square. However, $\displaystyle -1(y^2- y- 1/4)$ is NOT. And, of course, you don't want "$\displaystyle y^2$" inside the square. What you want is y^2- y+ 1/4- 1/4. The constant term on such a square is always positive. That is, you want $\displaystyle (x+ 1)^2+ (y- 1/2)^2+ z^2= 5/4$.

$\displaystyle (x^2+1)^2 + ( y^2 - 1/2)^2 + z^2 = 1 - 1/4 = 3/4$

r = $\displaystyle r^2 = 3/4$
r=$\displaystyle \sqrt {3/4}$

radius: $\displaystyle 2 \sqrt {3}$

Center: $\displaystyle (-1, + 1/2, 0)$

Am i correct?

hello

5. ## Re: Complete the square to find center and radius of the sphere

Thank you so much.