# Homework Help: Center of a Sphere in R^3

1. Sep 10, 2007

### RyanSchw

Center of a Sphere in R^3

I need to find the center and radius of a sphere given the equation:

$$x^2 - 4x + y^2 + z^2 = 0$$

I would like to hope it would be as easy as just adding 4x to both sides, but having a variable as the radius probably isn’t correct.

On the other hand, I have no idea how to complete the square when no other coefficients are present. Simply adding y and z coefficients only leads to variables I cannot get rid of on the right side of the equation.

Attempts I have thus far
$$(x-2)^2 + (y+2)^2 + (z+2)^2 = 12 + 4y + 4z$$
or
$$(x-2)^2+y^2+z^2=(\frac{4}{2})^2$$
I’m guessing I need to somehow get x^2 on the left and have the sphere centered about the origin.

Any help would be great, thanks.

2. Sep 10, 2007

### Kurdt

Staff Emeritus
You'll have to complete the square for the x variables then the answer should be straight forward. The cartesian equation of a square centered at (x0, y0, z0) is:

$$(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2= r^2$$

3. Sep 10, 2007

### Avodyne

If I give you the equation (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2, where a, b, c, and r are constants, does it describe a sphere? (Answer: yes.) What are the (x,y,z) coordinates of the center of the sphere? What is the radius?

4. Sep 10, 2007

### RyanSchw

So would it then be
$$(x-2)^2+(y-0)^2+(z-0)^2=(\frac{4}{2})^2$$?

That's about all I can think of =(

5. Sep 10, 2007

### Avodyne

OK, so what is the center? What is the radius?

6. Sep 11, 2007

### RyanSchw

If that's the equation for the sphere, the center would be (2,0,0) with r=2

7. Sep 11, 2007

### Kurdt

Staff Emeritus
Correct! Although I don't understand where you got 4/2 from?

8. Sep 11, 2007

### RyanSchw

The 4/2 was the factor that I added when I completed the square on both sides. I just left it that way because I wasnt sure where to go from there.

Thank you both for your help!

9. Sep 11, 2007

### Kurdt

Staff Emeritus
I only got a factor of 4. Never mind you have the correct answer anyway.