- #1
RyanSchw
- 36
- 0
Center of a Sphere in R^3
I need to find the center and radius of a sphere given the equation:
[tex]
x^2 - 4x + y^2 + z^2 = 0
[/tex]
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
[tex]
(x-2)^2 + (y+2)^2 + (z+2)^2 = 12 + 4y + 4z
[/tex]
or
[tex]
(x-2)^2+y^2+z^2=(\frac{4}{2})^2
[/tex]
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.
I need to find the center and radius of a sphere given the equation:
[tex]
x^2 - 4x + y^2 + z^2 = 0
[/tex]
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
[tex]
(x-2)^2 + (y+2)^2 + (z+2)^2 = 12 + 4y + 4z
[/tex]
or
[tex]
(x-2)^2+y^2+z^2=(\frac{4}{2})^2
[/tex]
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.