How to find these quantities so as to conform to these conditions?

Question

Suppose $a \in \mathbb{R}^k$, $b \in \mathbb{R}^k$. Then how to find $c \in \mathbb{R}^k$ and $r > 0$ such that the following holds?

For any $x \in \mathbb{R}^k$, we have $$|x-a| = 2 |x-b|$$ if and only if $$|x-c| = r. $$

Answer 1

Squaring both sides of the original equation and rewriting both sides as dot products gives $$(\mathbf{x} - \mathbf{a}) \cdot (\mathbf{x} - \mathbf{a}) = 4 (\mathbf{x} - \mathbf{b}) \cdot (\mathbf{x} - \mathbf{b}).$$ Expanding and rearranging gives $$3 \mathbf{x} \cdot \mathbf{x} + 2(\mathbf{a} - 4\mathbf{b}) \cdot \mathbf{x} + (4 \mathbf{b} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{a}) = 0.$$

Now, clearing the leading coefficient $3$ and completing the square gives:

$$\left(\mathbf{x} - \tfrac{1}{3}(4\mathbf{b} - \mathbf{a})\right) \cdot \left(\mathbf{x} - \tfrac{1}{3}(4\mathbf{b} - \mathbf{a})\right) + \tfrac{1}{3}(4 \mathbf{b} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{a}) - \tfrac{1}{9}(4\mathbf{b} - \mathbf{a}) \cdot (4\mathbf{b} - \mathbf{a}) = 0.$$ Can you take it from here?