I think this is a bit of an xy question. What you probably want to have on the long run is something that transforms you in a coordinate system the center of which is the center between two points, the x axis goes from the center to one of the two points, and the y axis is orthogonal and has length r. Then a circle in this coordinate system becomes your ellipse, ordinary arcs, elliptical arcs, and so on. Obviously, TikZ does that out of the box. To make things a bit more convenient I added a style, elli cs, that installs this coordinate system from these data. With this style, the graph becomes as simple as
\begin{scope}[elli cs={A={(P)},B={(Q)},r=9mm}]
\draw circle[radius=1];
\draw[dashed] (0,1) -- node[midway,fill=white]{$r$} (0,0) node[dot] {}
-- node[midway,fill=white]{$R$} (1,0);
\draw[blue,-{Stealth[bend]}] (-30:1) arc[start angle=-30,end angle=120,radius=1];
\end{scope}
As you see, for the arc there is no guess work required, the angles have a very intuitive interpretation in this coordinate system. You can do many further operations in this frame very conveniently, and also transform shapes according to it.
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{arrows.meta,bending,calc}
\begin{document}
\begin{tikzpicture}[dot/.style={circle,fill,inner sep=1pt},thick,
elli cs/.code={
\tikzset{ellipse through/.cd,#1}
\def\pv##1{\pgfkeysvalueof{/tikz/ellipse through/##1}}%
\edef\temp{\noexpand\tikzset{shift={($0.5*\pv{A}+0.5*\pv{B}$)},
x={($0.5*\pv{B}-0.5*\pv{A}$)},
y={($($\pv{A}!\pv{r}!90:\pv{B}$)-\pv{A}$)}
}}%
\temp
},
ellipse through/.cd,r/.initial=5mm,A/.initial={(-1,0)},
B/.initial={(1,0)}]
\path[nodes=dot] (0,0) node[label=below left:$P$] (P){}
(4,1.5) node[label=below right:$Q$] (Q){};
\begin{scope}[elli cs={A={(P)},B={(Q)},r=9mm}]
\draw circle[radius=1];
\draw[dashed] (0,1) -- node[midway,fill=white]{$r$} (0,0) node[dot] {}
-- node[midway,fill=white]{$R$} (1,0);
\draw[blue,-{Stealth[bend]}] (-30:1) arc[start angle=-30,end angle=120,radius=1];
\end{scope}
\end{tikzpicture}
\end{document}
