Euclidean space? (Rudin, page 23 Q 19) Question:
Suppose $\mathbf a\in R^k, \mathbf b\in R^k.$ Find $\mathbf c \in R^k$ and $r>0$ such that
$|\mathbf{x-a}| = 2|\mathbf{x-b}|$
if and only if $|\mathbf{x-c}| = r$
Solution: $3 \mathbf c = 4\mathbf b - \mathbf a, 3r = 2|\mathbf b-\mathbf a|$

I can see that the distance between points x and a is twice the distance of the points x and b. If r was the radius then we would have r = $(2/3)|\mathbf {b-a}|$ which gives $3r = 2|\mathbf {b-a}|$ as in the solution.
Now I'm not sure how to relate it to c. I'm thinking of doing
$|\mathbf{x - c}| = r$
$|\mathbf{x - c}| = (2/3)|\mathbf {b - a}|$
Am I on the right path? I don't know what to do next. I'm also not even sure if I'm supposed to assume r is the radius. If it weren't, then it seems anything could happen so r being the radius makes the most sense to me. Also, once I find c, is that sufficient for the if and only if condition?
Thank you.
 A: First, you can simplify some calculations by shifting coordinates. Let $x' = x-b, a'=a-b, b'=b-b = 0, c'=c-b$.
Then we want to find $c', r>0$ such that
$\|x'-a'\| = 2 \|x'\|$ iff $\|x'-c'\| = r$.
The equation $\|x'-a'\| = 2 \|x'\|$ expands to
$\|x'\|^2 + \|a'\|^2- 2 \langle a', x\rangle = 4 \|x'\|^2$, or
$\|x'\|^2 - 2 \langle -{1 \over 3} a', x\rangle = {1 \over 3} \|a'\|^2$.
The equation $\|x'-c'\| = r$ expands to
$\|x'\|^2  - 2 \langle c', x\rangle = r^2- \|c'\|^2$.
If we let $c' = -{1 \over 3} a'$ and $r = {2 \over 3} \|a'\|$ then these two equations are the same.
Computing $c=c'+b$ with $a = a'+b$ gives the desired result.
A: The reason to do special cases is that it tends to clear up problems with the notation, plus illustrates at least one example of the full problem.
In the plane, taking $a=(0,0)$ and $b=(6,0).$ The point you are writing $\mathbf{x} = (x,y).$ The relationship about double the distance becomes
$$ \color{blue}{ \sqrt{x^2 + y^2} = 2 \sqrt{(x-6)^2 + y^2}  }.  $$
Square,
$$ x^2 + y^2 = 4 \left( (x-6)^2 + y^2 \right), $$
$$ x^2 + y^2 = 4 \left( x^2 - 12 x + 36 + y^2 \right), $$
$$ x^2 + y^2 = 4  x^2 - 48 x + 144 + 4y^2 , $$
$$ 0 = 3  x^2 - 48 x + 144 + 3y^2 , $$
$$ 3  x^2 - 48 x + 144 + 3y^2 = 0 , $$
$$   x^2 - 16 x + 48 + y^2 = 0 , $$
$$  (x-8)^2 -64  + 48 + y^2 = 0 , $$
$$  (x-8)^2  + y^2 = 16 . $$
Circle, center $(8,0),$ radius $4.$
