# 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:

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

Attempt:

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

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

radius: $2 \sqrt {3}$

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

Am i correct?

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

You are not correct with:

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

but even if you were, this simplifies to:

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

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:

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

Attempt:

$(x^2 + 2x + 1) -1 (y^2 - y - 1/4) + 1/4 + z^2 = 0$
Yes, $x^2+ 2x+1= (x+ 1)^2$ is a perfect square. However, $-1(y^2- y- 1/4)$ is NOT. And, of course, you don't want " $y^2$" 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 $(x+ 1)^2+ (y- 1/2)^2+ z^2= 5/4$.

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

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

radius: $2 \sqrt {3}$

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

Am i correct?

hello

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

Thank you so much.