Show that the reflection of a disc through the origin is a disc This is a problem from the book "Basic Mathematics" by S.Lang (p.225, exercise 13b). It is similar to the one in my previous question, with the exception that we're considering a reflection instead of a translation.
Let $D(r, P)$ denote the disc of radius $r$ centered at $P$.
Show that the reflection of $D(r, P)$ through $O$ is a disc.
What is the center of this reflected disc?
Proof: Let $R_o$ denote the reflection through $O$. By definition
$$R_o(Q) = -Q$$
for some point $Q$.
Let $X$ be a point on $D(r, P)$. Then
$$d(X, P) \le r$$
For sake of argument, we choose $d(X, P) = s$, for some positive number $s \le r$.
$R_o$ is an isometry, and as such, for two distinct points $A, B$ we have
\begin{align}
d(A, B) & = d(R_o(A),R_o(B)) \\
& = d(-A, -B)
\end{align}
We know
$$R_o(P) = -P$$
Let $D(r, R_o(P))$ denote the disc of radius $r$ centered at $R_o(P)$. We wish to show that for any point $X$ on $D(r, P)$ there exists a unique point on $D(r, R_o(P))$ to which it gets mapped onto, under $R_o$, i.e., that the reflection $R_o$ of $D(r, P)$ is the disc of radius r, center $R_o(P) = -P$.
We have
\begin{align}
d(X, P) & = d(R_o(X), R_o(P)) &&\text{isometry} \\
& = \vert R_o(X) - R_o(P) \vert \\
& = \vert - X - (- P) \vert &&\text{substitution} \\
& = \vert P - X \vert \\
& = d(P, X) \\
& = d(X, P) \\
& = s \\
\end{align}
Whence, the image of $D(r, P)$ under $R_o$ is contained in $D(r, R_o(P))$.
Conversely, let $-X$ be a point on $D(r, R_o(P))$ such that
$$d(-X, R_o(P)) = s \tag{*} \label{*}$$
for some positive number $s \le r$.
We wish to show that every point on $D(r, R_o(P))$ can be expressed as the image of some point on $D(r, P)$, under $R_o$.
We have
\begin{align}
d(-X, R_o(P)) & = d(-X, -P) \\
& = d(R_o(X), R_o(P)) \\
& = d(X, P) \\
& = d(P, X) \\
& = s
\end{align}
Hence, the reflection of $D(r, P)$ through $O$ is the disc $D(r, R_o(P))$, of radius $r$, centered at $R_o(P) = -P$.$$\tag*{$\blacksquare$}$$
Question: This proof follows the same pattern as the proof in my previous question and that might explain some of it's ambiguity towards the end while trying to apply the same method I used for translations to reflections. 
Is the assumption that $\ref{*}-X$ is contained on $D(r, R_o(P))$ valid?
Thank you.
 A: Here's how I would prove it, if this helps:
We know that $R$ is an isometry, so $d(X, Y) = d(R(X), R(Y))$ for all points $X, Y$. We also know that $R$ is an involution, so $R(R(X)) = X$ for all points $X$. Using both these statements we get $d(R(X), Y) = d(R(R(X)), R(Y)) = d(X, R(Y))$ for all points $X, Y$.
We now have $$R(D(r,P)) = \{R(X) \mid d(X,P) < r\} = \{X \mid d(R(X), P) < r\} = \{X \mid d(X, R(P)) < r\} = D(r, R(P)).$$
In words:
Since a reflection is an isometry (preserves distances), the distance of the reflections of two points is the same as the distance of those two points. A reflection is also an involution, since reflecting a point twice yields the same point. The distance between $X$ and the reflection of $Y$ is, by isometry, the same as the distance between the reflection of $X$ and the reflection of the reflection of $Y$. That latter point is just $Y$ again.
Now the reflection of the disk around $P$ with radius $r$
is the reflection of all points who are closer to $P$ than $r$ 
is all points whose reflection is closer to $P$ than $r$ 
is all points who are closer to the reflection of $P$ than $r$ 
is the disk around the reflection of $P$ with radius $r$.
