Finding the radius of a sphere (3D coordinates)

[_dylanwhite]

So I came across an example problem in my textbook. I don't need to solve it, since it is solved for me, I just don't understand two of the steps it made.




In the actual math part, the four lines, I understand the move from the first to second line. It's just rearranging the equation to fit the sphere with a radius a at a specific center (shown). However, there must be some property of math that I absolutely missed and don't understand at all that led from line two to line three; from line three to four involves simplification and reverse foil.

So my questions are:
a) What concepts are needed to / how do you explain the movement from the second to third line in the picture example?
b) How do you infer the solution from the fourth line?

 

[HallsofIvy]

What you are missing is "completing the square". Given any numbers, x and a, $\displaystyle (x- a)^2= x^2- 2ax+ a^2$
so if you have something like $\displaystyle x^2- bx$, comparing that to $\displaystyle x^2- 2ax$ we see that we must have 2a= b or a= b/2. Then $\displaystyle a^2= b^2/4$ so that $\displaystyle x^2- bx= x^2- bx+ b^2/4- b^2/4f= (x- b/2)^2- b^2/4$.

Here, for example, you have $\displaystyle x^2+ 3x$ so that a= 3, a/2= 3/2, and $\displaystyle a^2/4= 9/4$. So we can write
$\displaystyle x^2+ 3x= x^2+ 3x+ 9/4- 9/4= (x+ 3/2)- 9/4$.

We have $\displaystyle z^2- 4z$ so that a= -4, a/2= -2, and $\displaystyle a^2= 4$. So we can write $\displaystyle z^2- 4z= z^2- 4z+ 4- 4= (z- 2)^2- 4$.

The "$\displaystyle y^2$" term is already a "perfect square": $\displaystyle y^2= (y- 0)^2+ 0$

We can apply that to this problem in either of two ways:
1) $\displaystyle x^2+ y^2+ z^2+ 3x- 4z+ 1= x^2+ 3x+ 9/4- 9/4+ y^2+ z^2- 4z+ 4- 4+ 1= 0$
$\displaystyle = (x+ 3/2)^2+ y^2+ (z- 2)^2- 9/4- 4+ 1= 0$
Now, add 9/4 and 4 and subtract 1 from both sides
$\displaystyle = (x+ 3/2)^2+ y^2+ (z- 2)^2= 9/4+ 4- 1= 9/4+ 16/4- 4/4= 21/4$

2) Subtract 1 from both sides to get $\displaystyle x^2+ 3x+ y^2+ z^2- 4z= -1$
Now add 9/4 and 4 to both sides.
$\displaystyle x^2+ 3x+ 9/4+ y^2+ z^2- 4z+ 4= 9/4+ 16/4- 1$
$\displaystyle (x+ 3/2)^2+ (y- 0)^2+ (z- 2)^2= 21/4$.

Compare that to $\displaystyle (x- x_0)^2+ (y- y_0)^2+ (z- z_0)^2= r^2$
for a sphere with center at $\displaystyle (x_0, y_0, z_0)$ and radius $\displaystyle r$.