Can someone please help me simplify the following formula for a sphere...?

Question

the given formula of the sphere is: x^2 + y^2 + z^2 + 2x - 2y - 4z - 3 = 0

I'm trying to get it in the form: (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2

but I'm not sure how.

Answer 1

The original equation=

x^2+2x+1-1+y^2-2y+1-1+z^2-4z+4-4-3=0

(x+1)^2+(y-1)^2+(z-2)^2=3^2

Answer 2

Compete the square (three times, once for each variable).

(x^2 + 2x + ?) + (y^2 -2y + ??) + (z^2 -4z + ???) - 3 - ? - ?? - ??? = 0

If you add a value (?) and subtract the same value (?), then you have not change the total sum on the left hand side.  That's because ? - ? = 0  (as long as both ? have the same value, of course).

Be careful with signs (square of negative value is positive, and if you get (x+a) then the centre of the sphere is at x= -a.

The first one is

x^2 + 2x + ?

The only "square" that fits in place of ? is +1.  It is the square of +1 and of -1.

Step 1, replace both ? by +1  (since the second one is subtracted, it becomes -1)

You now have

(x^2 + 2x + 1) + (y^2 -2y + ??) + (z^2 -4z + ???) - 3 - 1 - ?? - ??? = 0

? = + 1 is the square of +1 and of -1.  You must pick the corredt one: Look at the sign of the middle term in the square ( 2x).  The factor is positive (it is + 2x, not - 2x).

Therefore, the root to pick is +1.

The equation becomes:

(x+1)^2 + (y^2 -2y + ??) + (z^2 -4z + ???) - 4 - ?? - ??? = 0

Keep going