# Thread: equation of a sphere

1. ## equation of a sphere

i have to show this equation represents a sphere..

$x^2 + y^2 + z^2 = 8x - 4y - 4$

but i'm stuck on the z variable as it won't come down to an even square. it comes down to a (x+2)(x-2).

2. Originally Posted by greencheeseca
i have to show this equation represents a sphere..

$x^2 + y^2 + z^2 = 8x - 4y - 4$

but i'm stuck on the z variable as it won't come down to an even square. it comes down to a (x+2)(x-2).
Note that x^2 - 8x = (x - 4)^2 - 16 and y^2 + 4y = (y + 2)^2 - 4.

3. Originally Posted by mr fantastic
Note that x^2 - 8x = (x - 4)^2 - 16 and y^2 + 4y = (y + 2)^2 - 4.
ah, sorry i should have write (z+2)(z-2).. the other variables work out as even squares..

4. so, what can i do about the z variable? i can't get it to work out as an even square..

5. Originally Posted by greencheeseca
i have to show this equation represents a sphere..

$x^2 + y^2 + z^2 = 8x - 4y - 4$

but i'm stuck on the z variable as it won't come down to an even square. it comes down to a (x+2)(x-2).
1. The general equation of a sphere with the center $C(x_C, y_C, z_C)$ and the radius r is

$(x-x_C)^2+(y-y_C)^2+(z-z_C)^2=r^2$

2. Transform your equation until it has the form of the general equation:

$x^2 + y^2 + z^2 = 8x - 4y - 4$

Completing the square(s):

$x^2-8x + 16+y^2+4y+4 +z^2 = -4+16+4$

3. I'll leave the rest for you.

6. Originally Posted by greencheeseca
i have to show this equation represents a sphere..

$x^2 + y^2 + z^2 = 8x - 4y - 4$

but i'm stuck on the z variable as it won't come down to an even square. it comes down to a (x+2)(x-2).
Originally Posted by greencheeseca
so, what can i do about the z variable? i can't get it to work out as an even square..
The only term with z in it is z^2 and I don't see what your problem could be with it. I have already shown you how to deal with the x-terms and y-terms.

7. Originally Posted by greencheeseca
i have to show this equation represents a sphere..

$x^2 + y^2 + z^2 = 8x - 4y - 4$

but i'm stuck on the z variable as it won't come down to an even square. it comes down to a (x+2)(x-2).
You apparently are trying to do this as
$(x^2- 8x)+ (y^2+ 4y)+ (z^2- 4)= 0$
which is wrong for two reasons. First, moving that "- 4" to the other side of the equation makes it " $z^2+ 4$" which cannot be factored, not " $z^2- 4$".

But you don't want to do that to begin with!

Instead leave that constant, 4, on the right and do it as $(x^2- 8x)+ (y^2- 4y)+ (z^2)= -4$

Complete the square in x and y. The "z" term is already a perfect square! The numbers you need to add on the left to make the perfect squares, and, of course, add on the right, will give you a positive number on the right.