How do I verify the solution for this problem? (Advanced Calculus) I'm working on a problem in Rudin (Chapter 1, Exercise 1.19)...

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: $3 \mathbf c = 4\mathbf b - \mathbf a, 3r = 2|\mathbf b-\mathbf a|$

The solution is here: http://minds.wisconsin.edu/handle/1793/67009
Can somebody please explain to me why they squared both sides and the rest of the solution? I am confused on which equations they are squaring, how they squared it, and so on.
Thank you.

Working with the first equation, notice that:
\begin{align*}
&~|\mathbf{x-a}| = 2|\mathbf{x-b}| \\
&\iff |\mathbf{x-a}|^2 = 4|\mathbf{x-b}|^2 \qquad\text{since norms are nonnegative} \\
&\iff (\mathbf x - \mathbf a) \cdot (\mathbf x - \mathbf a) = 4[(\mathbf x - \mathbf b) \cdot (\mathbf x - \mathbf b)] \\
&\iff \mathbf x \cdot \mathbf x - \mathbf x \cdot \mathbf a - \mathbf a \cdot \mathbf x + \mathbf a \cdot \mathbf a = 4[\mathbf x \cdot \mathbf x - \mathbf x \cdot \mathbf b - \mathbf b \cdot \mathbf x + \mathbf b \cdot \mathbf b] \\
&\iff \mathbf{|x|}^2 - 2\mathbf{a \cdot  x + |a|}^2 = 4[\mathbf{|x|}^2 - 2\mathbf{b \cdot  x + |b|}^2]\\
&\iff \mathbf{|x|}^2 - 2\mathbf{a \cdot  x + |a|}^2 = 4\mathbf{|x|}^2 - 8\mathbf{b \cdot  x }+ 4\mathbf{|b|}^2 \\
&\iff 0 = 3\mathbf{|x|}^2 + 2\mathbf{a \cdot  x} - 8\mathbf{b \cdot  x} - |\mathbf a|^2+ {4|\mathbf b|}^2 \\
\end{align*}
Basically, we square both sides so that we can get rid of the norms and work with dot products instead.
