For the reflection $\phi$ in the sphere $S(a,r)$ it holds that $\phi(B^n) = B^n$ if and only if $\phi(a*) = 0$

Question

This is a question about Theorem 3.4.2 in "Geometry of discrete groups" from Beardon.
Let $\phi$ be the reflection in a sphere $S(a,r)$ for $a \in \hat{\mathbb{R}^n}$, $r \in \mathbb{R}$. Then TFAE:

1. $S(a,r)$ and $S^{n-1}$ are orthogonal
2. $\phi(a*) = 0$
3. $\phi(B^n) = B^n$.

where $a* := \frac{a}{{\lVert a \rVert}^2}$

I understand that 1 is equivalent to 2. But I don't understand why $3 \implies 2$. And in the proof of the implication $1+2 \implies 3 \:$ I do not understand the step

$\frac{r^2\cdot |x-a*|}{|x-a|\cdot|a*-a|} = \frac{|a|\cdot|x-a*|}{|x-a|}$

In the calculations. Thanks in advance for any help !

Answer 1

The equivalence of 1) <-> 3) can be found in Thm.4.4.6 of the book "Foundations of hyperbolic manifolds" by John G.Ratcliffe.