i have to show this equation represents a sphere..

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

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).

Printable View

- Oct 1st 2010, 05:19 AMgreencheesecaequation of a sphere
i have to show this equation represents a sphere..

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

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). - Oct 1st 2010, 05:20 AMmr fantastic
- Oct 1st 2010, 05:31 AMgreencheeseca
- Oct 1st 2010, 08:18 AMgreencheeseca
so, what can i do about the z variable? i can't get it to work out as an even square..

- Oct 1st 2010, 09:50 AMearboth
1. The general equation of a sphere with the center $\displaystyle C(x_C, y_C, z_C)$ and the radius r is

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

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

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

Completing the square(s):

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

3. I'll leave the rest for you. - Oct 1st 2010, 01:34 PMmr fantastic
- Oct 2nd 2010, 02:44 AMHallsofIvy
You apparently are trying to do this as

$\displaystyle (x^2- 8x)+ (y^2+ 4y)+ (z^2- 4)= 0$

which is wrong for two reasons. First, moving that "- 4" to the other side of the equation makes it "$\displaystyle z^2+ 4$" which cannot be factored, not "$\displaystyle z^2- 4$".

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

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

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.