# Thread: Completing the square of a sphere

1. ## Completing the square of a sphere

Hey,
I've got this equation and I was wondering what the steps are to get the answer becuase I got something different to the answer.

x^2 + y^2 + z^2 + 4x 6y + 2z + 6 = 0

Complete the square to find the origin and the radius

I got the origin all right but my radius came out to be sqrt(20) and the answer says it's sqrt(8).
If someone could explain that would be great.

2. Originally Posted by Evales
Hey,
I've got this equation and I was wondering what the steps are to get the answer becuase I got something different to the answer.

x^2 + y^2 + z^2 + 4x 6y + 2z + 6 = 0

Complete the square to find the origin and the radius

I got the origin all right but my radius came out to be sqrt(20) and the answer says it's sqrt(8).
If someone could explain that would be great.
you mean "center", not "origin".

bring the 6 to the right of the equation to get -6.

to complete the square for the x's, you need to add 4 to both sides

to complete the square for the y's, you need to add 9 to both sides

to complete the square for the z's, you need to add 1 to both sides

thus the right side becomes -6 + 4 + 9 + 1 = 8 which is the square of the radius. hence the radius is $\displaystyle \sqrt 8$

(Look up how to complete the square)

3. Originally Posted by Evales
Hey,
I've got this equation and I was wondering what the steps are to get the answer becuase I got something different to the answer.

x^2 + y^2 + z^2 + 4x 6y + 2z + 6 = 0

Complete the square to find the origin and the radius

I got the origin all right but my radius came out to be sqrt(20) and the answer says it's sqrt(8).
If someone could explain that would be great.
It helps if you write all $\displaystyle x$ terms together, all $\displaystyle y$ terms together, and all $\displaystyle z$ terms together.

So

$\displaystyle x^2 + 4x + y^2 - 6y + z^2 + 2z = -6$.

Now each pair of terms represents the first two terms of a quadratic.

To find the third term, you need to halve the coefficient of the second and then square it. Then you add this term to both sides of the equation.

So

$\displaystyle x^2 + 4x + 2^2 + y^2 - 6y + (-3)^2 + z^2 + 2z + 1^2 = -6 + 2^2 + (-3)^2 + 1^2$.

Now each trio of terms is a perfect square. So you can rewrite it as

$\displaystyle (x + 2)^2 + (y - 3)^2 + (z + 1)^2 = 8$.

And now, writing in standard form

$\displaystyle [x - (-2)]^2 + (y - 3)^2 + [z - (-1)]^2 = \left(\sqrt{8}\right)^2$

$\displaystyle [x - (-2)]^2 + (y - 3)^2 + [z - (-1)]^2 = \left(2\sqrt{2}\right)^2$.

So the centre is $\displaystyle (x, y, z) = (-2, 3, -1)$ and the radius is $\displaystyle r = 2\sqrt{2}$.

4. Thanks guys, I was only adding it to one side (duh) haha. Brain is out of practice.