Prob. 19, Chap. 1, in Baby Rudin: For what $\mathbf{c}$ and $r > 0$ does this equivalence hold?

Question

Here's Prob. 19 in Chap. 1 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $\mathbf{a} \in \mathbb{R}^k$, $\mathbf{b} \in \mathbb{R}^k$. Find $\mathbf{c} \in \mathbb{R}^k$ and $r > 0$ such that, for all $\mathbf{x} \in \mathbb{R}^k$, we have $$\vert \mathbf{x} - \mathbf{a} \vert = 2 \vert \mathbf{x} - \mathbf{b} \vert$$ if and only if $$\vert \mathbf{x} - \mathbf{c} \vert = r.$$

Although Rudin has given a solution, namely $3\mathbf{c} = 4 \mathbf{b} - \mathbf{a}$, $3r = 2\vert \mathbf{b} - \mathbf{a} \vert$, I'm wondering how he's obtained it.

How to attack this type of a problem?

Is this problem part of the exercises by some well-thought-out design? I mean is it going to be used later on in the book? Or, is it just to give the reader some practice?

Answer 1

$$\begin{align} |\mathbf x-\mathbf a|=2|\mathbf x-\mathbf b| & \iff \left[\sum_{n=1}^k(x_n-a_n)^2\right]^{1/2}=2\left[\sum_{n=1}^k(x_n-b_n)^2\right]^{1/2}\\ & \iff \sum_{n=1}^k(x_n-a_n)^2=4\sum_{n=1}^k(x_n-b_n)^2\\ & \iff \sum_{n=1}^k(x_n^2-2a_nx_n+a_n^2)=\sum_{n=1}^k(4x_n^2-8b_nx_n+4b_n^2)\\ & \iff \sum_{n=1}^k[3x_n^2+(2a_n-8b_n)x_n]=\sum_{k=1}^n(a_n^2-4b_n^2)\\ & \iff \sum_{n=1}^k\left[x_n^2+\frac13(2a_n-8b_n)x_n\right]=\frac13\sum_{n=1}^k(a_n^2-4b_n^2)\\ & \iff \sum_{n=1}^k\left[x_n-\frac13(4b_n-a_n)\right]^2=\frac19\sum_{n=1}^k(4a_n^2-8a_nb_n+4b_n^2)\\ & \iff \left|\mathbf x-\frac13(4\mathbf b-\mathbf a)\right|=\frac23|\mathbf b-\mathbf a| \end{align}$$

So, $3\mathbf c=4\mathbf b-\mathbf a$ and $3r=2|\mathbf b-\mathbf a|$.

This problem concerns circles of Apollonius. See here: https://en.wikipedia.org/wiki/Circles_of_Apollonius

Answer 2

First a comment. What this problem really says is that the subset of $\mathbb{R}^k$ defined by the equation $|x - a| = 2|x - b|$ is a sphere. So there is really not going to be any choice in determining $c$ and $r$. The point $c$ will need to be the centre of the sphere and $r$ its radius.

I imagine the main point of this exercise is to give readers practice with dot products, particularly the relation $|x|^2 = x \cdot x$.

To prove the result, write the following succession of equivalent equations, calculations that amount to completing the square with dot products. $$ \begin{align*} 2|x - b| &= |x - a| \\ 4|x-b|^2 &= |x - a|^2 \\ 4 (x - b) \cdot (x - b) &= (x - a) \cdot (x - a) \\ 4 x \cdot x - 8 x \cdot b + 4 b \cdot b &= x \cdot x - 2 x \cdot a + a \cdot a \\ 3 x \cdot x - 8 x \cdot b + 2 x \cdot a &= a \cdot a - 4 b \cdot b \\ x \cdot x - 2 x \cdot \left( \frac{4}{3}b - \frac{1}{3} a\right) &= \frac{1}{3}a \cdot a - \frac{4}{3}b \cdot b \\ x \cdot x - 2 x \cdot \left( \frac{4}{3}b - \frac{1}{3} a\right) +\left( \frac{4}{3}b - \frac{1}{3} a\right) \cdot \left( \frac{4}{3}b - \frac{1}{3} a\right) &= \frac{1}{3}a \cdot a - \frac{4}{3}b \cdot b + \left( \frac{4}{3}b - \frac{1}{3} a\right) \cdot \left( \frac{4}{3}b - \frac{1}{3} a\right)\\ (x - c) \cdot (x -c) &= r^2 \\ |x - c| &= r \end{align*} $$ where $c = \frac{4}{3}b - \frac{1}{3} a$ and $r^2 = \frac{4}{9} a \cdot a - \frac{8}{9} a \cdot b + \frac{4}{9} b \cdot b = \frac{4}{9}(a - b) \cdot (a - b)$, so $r = \frac{2}{3}|a - b|$.

Technically, the exercise is incorrect when $a = b$.