Important additional assumption: we want to measure the "small" arcs with endpoints $PAB(r)$ and $PAC(r)$, and
the half-line originating from $O$ and going through $A$ must intersect those arcs.
If you need something more general, the approach below will fail, but it can be adapted.
The arc length in general is "radius" $\times$ "angular width (in radian)". We already know the radius is $r_1+r$, so we need to evaluate the angle. With my additional assumption on the half-line from $O$ through $A$, it suffices to evaluate the two angles $\widehat{PAB\ O\ A}$ and
$\widehat{A\ O\ PAC}$.
Below I go over how to obtain $\widehat{PAB\ O\ A}$, the other angle can be obtained in a similar way.
Let $x$ be the distance between $A$ and $PAB$, $\alpha$ the angle $\widehat{OAB}$ and $\theta$ our angle of interest $\widehat{PAB\ O\ A}$.
Angle $\alpha$ is a parameter of the problem, $x$ and $\theta$ vary with $r$.
We only need to know the cosine of $\alpha$, which can be obtained by the relation
$$
(r_1+r_2)^2=AB^2+r_1^2-2ABr_1\cos\alpha
$$
Next we want $x$:
\begin{align*}
&(r_1+r)^2=x^2+r_1^2-2xr_1\cos\alpha=(x-r_1\cos\alpha)^2-r_1^2\cos^2\alpha+r_1^2\\
&\iff\quad
(x-r_1\cos\alpha)^2=r_1^2+r^2+2r_1r+r_1^2\cos^2\alpha-r_1^2\\
&\iff\quad
x=r_1\cos\alpha\pm\sqrt{r^2+2r_1r+r_1^2\cos^2\alpha}
\end{align*}
Now because $x\ge 0$ and $\sqrt{r^2+2r_1r+r_1^2\cos^2\alpha}\ge\lvert r_1\cos\alpha\rvert$ we actually have
$x=r_1\cos\alpha + \sqrt{r^2+2r_1r+r_1^2\cos^2\alpha}$.
Next is $\theta$:
$$
x^2=(r_1+r)^2+r_1^2-2(r_1+r)r_1\cos\theta
$$
So if you put everything together you easily get $\cos\theta$ with respect to the various parameters.
Now the problem is that the identity $\theta=\arccos\big(\cos\theta\big)$
is only true when $0\le\theta\le\pi$. That's where my two additional assumptions come into play, since they guarantee $0\le\theta\le\pi$.
