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](https://en.wikipedia.org/wiki/Circular_segment) on that larger circle:

[![circles image][1]][1]

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:

$$\begin{align}
-2r(1+√3) + 5+2√3 + 1 & = 0 \\
-2r(1+√3) + 6+2√3 & = 0 \\
-r(1+√3) + 3+√3 & = 0 \\
r(1+√3) - (3+√3) & = 0 \\
r(1+√3) & = (3+√3) \\
r & = (√3+3)/(1+√3) \\
  & = √3(√3+3)/√3(1+√3) \\
  & = √3(√3+3)/(√3+3) \\
  & = √3
\end{align}$$

  [1]: https://i.sstatic.net/dVMCE.png