Since the area of an equilateral triangle with the side length $a$ is $\frac{\sqrt 3}{4}a^2,$ we have
$$4\sqrt 3=\frac{\sqrt 3}{4}a^2\Rightarrow a=4.$$
Then, letting $R$ be the radius of the circumcircle, we have, by the law of sines,
$$\frac{4}{\sin(60^\circ)}=2R\Rightarrow R=\frac{4}{3}\sqrt 3.$$
Suggested approach: Prove that $TF \times TX = TB \times TA$, where lines $DE$ and $AB$ intersect at point $T$. (In particular, independent of the $P,Q$, as suggested by the problem.)
Corollary: It follows that $TF \times TX = TB \times TA = TP \times TQ$ and thus $F,X,P,Q$ are concyclic as desired.
Proof of approach: One way to prove the equation is by side length chasing. Apply Menelaus on triangle $ABC$ to transversal $TDE$ to obtain $TA/TB$ and hence $TA, TB$. Then we can find $TF, TX$ and multiply it out.
Details of side length chasing
$\frac{AT}{TB} \times \frac{BD}{DC} \times \frac{CE}{EA} = 1 $
$ \frac{AT} {TB} = \frac{ EA}{BD} = \frac{c+b-a}{c+a-b}$
$AT - TB = c \Rightarrow AT = \frac{ c (c+b-a) } { 2(b-a) } , TB = \frac{ c(c+a - b ) } { 2(b-a)}$
$TF = TB + BF = \frac{ c(c+a - b ) } { 2(b-a)} + \frac{c+a - b}{2} = \frac{ ((c-b+a)(c+a-b) } { 2(b-a)} $
$TX = TB + BX = \frac{ c(c+a - b ) } { 2(b-a)} + \frac{c}{2} = \frac{ c(c)}{ 2(b-a)}$
Now, we multiply these terms to show that $TA \times TB = \frac{ c^2 (c+b-a)(c+a-b) } { 4(b-a)^2} = TX \times TF$
Additional observations, which I couldn't use directly
$A, F, T, B$ are harmonic conjugates. This can be shown either from
1) $ \frac{ TA}{TB} = \frac{EA}{BD} = \frac{AF}{FB}$, or also from
2) Lines $AD, BE, CF$ are concurrent (at the Gergonne point)
Best Answer
do you mean the formula of Euler? here it is https://proofwiki.org/wiki/Euler_Triangle_Formula