Defining a sphere

Question

Let $c$ be a number, and let $a$ and $b$ be vectors in $R^3$. Let x = $(x, y, z)$. Show that the equation (x$-a)\cdot($x$-b)=c^2$ defines a sphere with centre whose position vector is $1/2(a+b)$ and radius $[c^2+|1/2(a-b)|^2]^{1/2}$.

My working:

Letting x = $(x, y, z)$, $b = (b_1, b_2, b_3)$, $a = (a_1, a_2, a_3)$,

(x$-a)\cdot($x$-b)=c^2$

$|x|^2 - b \cdot x - a \cdot x+a \cdot b=c^2$

$x^2+y^2+z^2-(b_1x+b_2y+b_3z)-(a_1x+a_2y+a_3z)+(b_1a_1+b_2a_2+b_3a_3)-c^2=0$

$x^2+y^2+z^2+b_1a_1(1-x)+b_2a_2(1-y)+b_3a_3(1-z)-c^2=0$

After simplification, I get

$(x-\frac{b_1a_1}{2})^2+(y-\frac{b_2a_2}{2})^2+(z-\frac{b_3a_3}{2})^2=c^2-a \cdot b+\frac{(b_1)^2(a_1)^2}{4}+\frac{(b_2)^2(a_2)^2}{4}+\frac{(b_3)^2(a_3)^2}{4}$

From here, I am unsure of how to further simpify to prove the given position vector and radius, may I have some help please?

Answer 1

You can do it like this. For $r \in \mathbb{R}^3$ let us write $r^2 = r \cdot r = \lvert r \rvert^2$. Moreover, let us write $p = (x,y,z)$ instead of x. Then

$$c^2 = (p -a) \cdot (p - b) = p^2 -(a + b) \cdot p + a \cdot b \\ = p^2 -(a + b) \cdot p + \left((-\frac{1}{2})(a + b)\right)^2 - \left((-\frac{1}{2})(a + b)\right)^2 + a \cdot b \\ = \left(p -\frac{1}{2}(a + b) \right)^2 - \left(\frac{1}{2}(a - b) \right)^2 \\ $$ and we are done.

Answer 2

Probably simpler to work backwards. If $\vec x$ lies on a sphere with centre $\vec{O}$ and radius $r$ then

$|\vec x - \vec O|^2 = r^2 \\ \Rightarrow (\vec x - \vec O) \cdot (\vec x - \vec O) = r^2 \\ \Rightarrow \vec x \cdot \vec x -2\vec x \cdot \vec O = r^2 - \vec O \cdot \vec O \qquad (1)$

Expanding $(\vec x- \vec a) \cdot (\vec x-\vec b) = c^2$ we have

$\vec x \cdot \vec x - \vec x \cdot (\vec a + \vec b) = c^2 - \vec a \cdot \vec b \qquad (2)$

Comparing $(1)$ and $(2)$ we have

$\vec O = \frac{\vec a + \vec b}{2}$

$r^2 = c^2 - \vec a \cdot \vec b + \vec O \cdot \vec O = c^2 -\frac{\vec a \cdot \vec a - 2 \vec a \cdot \vec b + \vec a \cdot \vec a}{4} = c^2 - \left | \frac{\vec a - \vec b}{2}\right |^2$