Problems like this make me want to look for general principles instead of bogging-down in messy specifics. In this case, we have a triangle $\triangle ABC$, say, with circumcenter $O$, and circular-segment-centroids $A'$, $B'$, $C'$ (OP's $A_1$, $B_1$, $C_1$), with $\overline{OA'}$, $\overline{OB'}$, $\overline{OC'}$ bisecting $\angle BOC$, $\angle COA$, $\angle AOB$, respectively; a little calculus tells us the centroids' distances from the circumcenter. (We'll get to that later.)
Stepping back, we see that we have six points $A$, $B$, $C$, $A'$, $B'$, $C'$ arranged about a common center, $O$, and that we know the distances of these point from $O$ and angles determined by these points and $O$. We can establish a condition on those lengths and angles that guarantees the concurrence of $\overleftrightarrow{AA'}$, $\overleftrightarrow{BB'}$, $\overleftrightarrow{CC'}$. So let's do that.
Define
$$
a := |OA| \quad b := |OB| \quad c := |OC| \quad
a' := |OA'| \quad b' := |OB'| \quad c' := |OC'|
$$
(Note that we're generalizing beyond $O$ being the circumcenter, which would required $a=b=c$. We're also generalizing beyond, say, $\overline{OA'}$ bisecting $\angle BOC$, etc; the general rule turns out to be pretty nice without these assumptions.) We'll also use various angles, $\angle XOY$, taken to be oriented "from" $X$ "to" $Y$; this allows us to write $\angle XOY+\angle YOZ=\angle XOZ$ and $\angle XOY=-\angle YOX$.
Now, let's coordinatize. Abusing notation to define $\operatorname{cis}\theta := (\cos\theta, \sin\theta)$ we can take
$$\begin{align}
A &:= a \operatorname{cis}0
&& A' := a'\operatorname{cis}\angle AOA'\\
B &:= b \operatorname{cis}\angle AOB
&& B' := b'\operatorname{cis}\angle AOB' \\
C &:= c\operatorname{cis}\angle AOC
&& C' := c'\operatorname{cis}\angle AOC'
\end{align}$$
From here, the process is straightforward, if tedious. (It helps to have a computer algebra system to crunch the symbols.) We determine the equations of the lines $\overleftrightarrow{AA'}$, $\overleftrightarrow{BB'}$, $\overleftrightarrow{CC'}$, find the intersection of any two, and substitute the intersection into the third. When the dust settles (and barring degeneracies), we get a relation that we can express thusly:
$$\begin{align}
0 &= \phantom{+}a a' \sin\angle AOA'\; \left(
b c \sin\angle BOC
+c b' \sin\angle COB'
+b' c' \sin\angle B'OC'
+c' b \sin\angle C'OB
\right) \\[4pt]
&\phantom{=} +b b'\sin\angle BOB'\; \left(
c a \sin\angle COA
+a c' \sin\angle AOC'
+c' a' \sin\angle C'OA'
+a' c \sin\angle A'OC
\right) \\[4pt]
&\phantom{=}+ c c' \sin\angle COC'\; \left(
a b \sin\angle AOB
+b a' \sin\angle BOA'
+a' b' \sin\angle A'OB'
+b' a \sin\angle B'OA
\right)
\end{align} \tag{$\star$}$$
This may seem a little daunting at first glance, but, glancing again, we notice that every "$\sin\angle XOY$" is multiplied by the corresponding lengths "$x$" and "$y$"; conveniently, each such product is thus twice the (signed) area $|\triangle XOY|$, so that we can write
$$\begin{align}
0 &= \phantom{+}|\triangle AOA'|\; \left(
|\triangle BOC|+|\triangle COB'|+|\triangle B'OC'|+|\triangle C'OB|
\right) \\[4pt]
&\phantom{=} +|\triangle BOB'|\; \left(
|\triangle COA|+|\triangle AOC'|+|\triangle C'OA'|+|\triangle A'OC|
\right) \\[4pt]
&\phantom{=}+ |\triangle COC'|\; \left(
|\triangle AOB|+|\triangle BOA'|+|\triangle A'OB'|+|\triangle B'OA|
\right)
\end{align} \tag{$\star\star$}$$
Even better, each long factor is the sum of the (signed) areas of adjacent triangles that form a quadrilateral; so each factor gives the (signed) area of that quadrilateral. (This interpretation is a bit nuanced in cases where such a quadrilateral is self-intersecting. Be that as it may ...) This gives us this streamlined expression:
$$
|\triangle AOA'|\;|\square BCB'C'|
+|\triangle BOB'|\;|\square CAC'A'|
+|\triangle COC'|\;|\square ABA'B'|
=0
\tag{$\star\star\star$}$$
Pretty nifty! $\square$
Now that we have generalized the problem, let's work our way towards the specifics of OP's ostensible concurrence.
We consider $\triangle ABC$ with interior angles $\alpha := \angle A$, $\beta := \angle B$, $\gamma := \angle C$. Taking $O$ to be the circumcenter and $r$ the circumradius, we have
$$a=b=c=r \qquad \angle BOC = 2\alpha \quad \angle COA = 2\beta \quad \angle AOB = 2\gamma $$
With $A'$, $B'$, $C'$ along the bisectors of $\angle BOC$, $\angle COA$, $\angle AOB$, we have
$$\angle BOA' = \angle A'OC=\alpha \qquad \angle COB'=\angle B'OA=\beta \qquad
\angle AOC'=\angle C'OB=\gamma$$
$$\angle AOA' = 2\gamma+\alpha=\pi-(\beta-\gamma) \qquad
\angle BOB' = \pi-(\gamma-\alpha) \qquad
\angle COC' = \pi-(\alpha-\beta)$$
Falling back to version $(\star)$ of our concurrence condition, we have
$$\begin{align}
0 &= \phantom{+}r a' \sin(\beta-\gamma)\; \left(
r^2 \sin2\alpha
+r b' \sin\beta
+b' c' \sin(\beta+\gamma)
+c' r \sin\gamma
\right) \\[4pt]
&\phantom{=} +r b'\sin(\gamma-\alpha)\; \left(
r^2 \sin2\beta
+r c' \sin\gamma
+c' a' \sin(\gamma+\alpha)
+a' r \sin\alpha
\right) \\[4pt]
&\phantom{=}+ r c' \sin(\alpha-\beta)\; \left(
r^2 \sin2\gamma
+r a' \sin\alpha
+a' b' \sin(\alpha+\beta)
+b' r \sin\beta
\right)
\end{align} \tag{1}$$
Since $\alpha+\beta+\gamma=\pi$ and $r\neq 0$, this simplifies to
$$\begin{align}
0 &= \phantom{+}(r a' - b' c') \sin2\alpha \sin(\beta - \gamma) \\
&\phantom{=}+(r b' - c' a') \sin2\beta \sin(\gamma - \alpha) \\
&\phantom{=}+(r c' - a' b') \sin2\gamma \sin(\alpha - \beta)
\end{align} \tag2$$
Note that $(2)$ holds for $A'$, $B'$, $C'$ anywhere along the perpendicular bisectors, so it's still a bit of a generalized result. For OP's circular-segment-centroids, we consult Wikipedia's "List of Centroids" to remind ourselves that
$$a' = \frac{4r\sin^3\alpha}{3(2\alpha-\sin2\alpha)} \qquad
b' = \frac{4r\sin^3\beta}{3(2\beta-\sin2\beta)} \qquad
c' = \frac{4r\sin^3\gamma}{3(2\gamma-\sin2\gamma)} \tag{3}$$
Perhaps-unsurprisingly, upon substituting the values from $(3)$ into $(2)$, the mix of "raw and trigged" angles doesn't simply vanish. For the sake of completeness, here's a version of the resulting concurrence condition
$$\begin{align}
&\phantom{=+\,}
3 \sin^3\alpha \sin(\beta-\gamma) (\alpha \sin2\beta\sin2\gamma + 2\beta\gamma \sin2\alpha) \\
&\phantom{=}
+3 \sin^3\beta \sin(\gamma-\alpha) (\beta \sin2\gamma\sin2\alpha + 2\gamma\alpha \sin2\beta) \\
&\phantom{=}
+3 \sin^3\gamma \sin(\alpha-\beta) (\gamma \sin2\alpha\sin2\beta + 2\alpha\beta \sin2\gamma) \\[6pt]
&= 8 \sin\alpha \sin\beta \sin\gamma \left(\begin{array}{l}
\phantom{+}
\alpha \cos\alpha \sin^2\beta \sin^2\gamma \sin(\beta-\gamma) \\
+ \beta \cos\beta \sin^2\gamma \sin^2\alpha \sin(\gamma-\alpha) \\
+ \gamma \cos\gamma \sin^2\alpha \sin^2\beta \sin(\alpha-\beta) \\
+ \sin\alpha \sin\beta \sin\gamma \sin(\beta-\gamma) \sin(\gamma-\alpha) \sin(\alpha-\beta)\end{array}\right)
\end{align}\tag{4}$$
OP's alternative construction, taking $A'$, $B'$, $C'$ to be the centroids of the "other" circular segments, requires the substitutions $\alpha\to\pi-\alpha=\beta+\gamma$, $\beta\to\gamma+\alpha$, $\gamma\to\alpha+\beta$ in $(3)$, along with changing the sign of each of $a'$, $b'$, $c'$ because each centroid lies on the "other side" of $O$. These adjustments cause some minor sign changes in $(4)$, but also introduce more-complicated "raw" angle expressions. The result doesn't appreciably simplify, so I won't bother TeX-ing it up.
Best Answer
Given that $|PQ|=|PR|=5,\ |QR|=6$, the area, the height and the circumradius of $\triangle PQR$ are $S=12$, $|PF|=4$ and $R_0=\tfrac{25}8$, respectively. Let $\angle PQR=\alpha$, $\angle FOE=\phi$.
Assuming that the center of the circle $O_t\in QR$, we must have $|DQ_t|=|EQ_t|=r$.
\begin{align} \sin\alpha&=\frac{|PF|}{|PQ|} =\frac45 ,\\ |OF|&=|PF|-R_0=\tfrac78 \tag{1}\label{1} . \end{align}
We have two conditions for $r$, $\phi$:
\begin{align} |QO_t|+|FO_t|&=\tfrac12\,|QR| \tag{2}\label{2} ,\\ \frac r{\sin\alpha} + (R_0-r)\sin\phi &=3 \tag{3}\label{3} ,\\ (R_0-r)\cos\phi&=|OF| \tag{4}\label{4} . \end{align}
Excluding $\phi$ from \eqref{3},\eqref{4} and using known values, we get
\begin{align} r&=\frac{20}9 . \end{align}