In $\mathbb R^n$ prove $\{ y : |y| = R\} = \{ y : |y-x| = \frac{|x|}{R } | y - \bar{x} | \} $ where $|x| < R$ and $\bar{x}$ is the reflection of $x$

Question

In $\mathbb R^n$, let $x \in B_R(0)$. prove that $\{ y \in \mathbb R^n : |y| = R \} = \{ y \in \mathbb R^n : |y-x| = \frac{|x|}{R}|y - \bar{x}| \}$ where $\bar{x} : = \frac{R^2}{|x|^2}x$ is the reflection of $x$ along the circle of radius $R$ centered at the origin.

This comes in the course of proving the Poisson Integral Formula in $\mathbb R^n$. I do get why the formula for $\bar{x}$ represents the point as the reflection of the point along the circle. I am not sure, however, why the two sets must be equal in algebra. More importantly, I do not get why anyone would intuitively get that the two sets are equal and/or be able to visualize the two sets as being equal. Could anyone explain in plain words why the two sets must be equal? If nothing else, I would appreciate straightforward calculation showing the two sets are equal, since I somehow cannot get the algebra. I am thinking that perhaps my professor may have misstated this on the board, God bless him.

Could anyone also refer to me a link for the proof of the multidimensional Poisson Integral Formula?

Answer 1

Presumably $x\ne0$. Let $v=y/R$ and $au=x/R$, where $0<a=|x|/R<1$ and $u=x/|x|$ is a unit vector. Then \begin{align*}&|y-x|=\frac{|x|}{R}|y-\overline{x}|\\ &\Leftrightarrow |v-au|=|av-u|\\ &\Leftrightarrow 0=|av-u|^2-|v-au|^2=(1-a^2)(1-|v|^2)\\ &\Leftrightarrow |v|=1\\ &\Leftrightarrow |y|=R.\end{align*}

Answer 2

Suppose $|y|=R$. Using the fact that $|a-b|^{2}=|a|^{2}+|b|^{2}-2\langle x, y \rangle$ we get (after some simplification) $\frac {|x|^{2}} {R^{2}} |y-\overline x|^{2}=R^{2}+|x|^{2}-2\langle x, y \rangle=|y-x|^{2}$. This proves LHS $\subseteq $ RHS. I will leave the other inclusion to you.

Answer 3

Here is some intuition: This fact is related to Apollonian circles. Consider the segment joining $x$ and $\bar x$. By definition of reflection, the segment is radial. The set on the right hand side of your set equality is the set of points such that the ratio $|y-x|/|y-\bar x|$ is constant. On each two-dimensional plane containing the segment $x\bar x$, this is an Apollonian circle, centered at the origin and with radius $R$. The union of those circles forms the sphere of radius $R$ which is your left-hand side.

To see why the circles have radius $R$ centered at the origin, it is enough to check that the points $y=\pm Rx/|x|$ belong to all of them.