Slightly late to the party. This finds $r$ of the larger circle (where the smaller circle is a unit radius) using $FHD$ as a circular segment on that larger circle:
Some generalities, not all of which are used:
$$\begin{align} AB & = BC = BJ = DE = EF = EK = EG = GB = 1 \\ AC & = EB = 2 \end{align}$$
Use Pythagoras to establish relative location of $E$, $B$ and $K$:
$$\begin{align} ∠EKB & = π/2 \\ EB^2 & = EK^2 + BK^2 \\ BK & = √(EB^2 - EK^2) \\ & = √(4 - 1) = √3 \end{align}$$
Cartesian coordinates for all the points identified except $D$, $F$, $L$ as they are more complicated:
$$\begin{align} A & = (2r, 0) \\ B & = (2r-1, 0) \\ C & = (2r-2, 0) \\ E & = (2r-1-√3, 1) \\ G & = (B + E)/2 \\ & = 1/2(2r-1 + 2r-1-√3, 1) \\ & = 1/2(4r-2-√3, 1) \\ & = (2r-1-√3/2, 1/2) \\ H & = (r, 0) \\ J & = (2r-1, 1) \\ K & = (2r-1-√3, 0) \end{align}$$
Establish $EH$ using Pythagoras, in order to get the apothem of the segment:
$$\begin{align} HK & = H - K \\ & = r - (2r-1-√3) \\ & = r - 2r + 1 + √3 \\ & = 1 + √3 - r \\ HK^2 & = (1 + √3 - r)^2 \\ & = (1 + √3)^2 - 2r(1 + √3) + r^2 \\ & = 1 + 2√3 + 3 - 2r(1 + √3) + r^2 \\ & = 4 + 2√3 - 2r(1 + √3) + r^2 \\ EH^2 & = HK^2 + EK^2 \\ & = 4 + 2√3 - 2r(1 + √3) + r^2 + 1 \\ & = 5 + 2√3 - 2r(1 + √3) + r^2 \\ & = r^2 - 2r(1+√3) + 5+2√3 \\ EH = & √(r^2 - 2r(1+√3) + 5+2√3) \end{align}$$
$r$ in terms of $h$ (the sagitta, $EL$) and $c$ (the chord length, $DF$) via the intersecting chord theorem:
$$\begin{align} r & = c^2/8h + h/2 = HL \\ c & = DF = 2 \\ h & = EL = HL - EH \\ & = r - √(r^2 - 2r(1+√3) + 5+2√3) \\ r & = 4/8h + h/2 = 1/2h + h/2 \\ 2hr & = h^2 + 1 \\ h^2 - 2hr + 1 & = 0 \end{align}$$
Expand out $h^2$ and $2hr$ because then it is very clean to complete the equation above:
$$\begin{align} h^2 & = r^2 - 2r√(r^2 - 2r(1+√3) + 5+2√3) + r^2 - 2r(1+√3) + 5+2√3 \\ & = 2r^2 - 2r√(r^2 - 2r(1+√3) + 5+2√3) - 2r(1+√3) + 5+2√3 \\ 2hr & = 2r^2 - 2r√(r^2 - 2r(1+√3) + 5+2√3) \end{align}$$
Then we have:
$$\begin{align} -2r(1+√3) + 5+2√3 + 1 & = 0 \\ -2r(1+√3) + 6+2√3 & = 0 \\ -r(1+√3) + 3+√3 & = 0 \\ r(1+√3) - (3+√3) & = 0 \\ r(1+√3) & = (3+√3) \\ r & = (√3+3)/(1+√3) \\ & = √3(√3+3)/√3(1+√3) \\ & = √3(√3+3)/(√3+3) \\ & = √3 \end{align}$$