Return to Answer

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:

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

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:

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

Feeding $r$ back into $EH$:

$$\begin{align} r & = √3 \\ EH & = √(r^2 − 2r(1+√3) + 5 + 2√3) \\ & = √(3 − 2√3(1+√3) + 5 + 2√3) \\ & = √(8 − (2√3+6) + 2√3) \\ & = √(8 − 2√3 - 6 + 2√3) \\ & = √(8 - 6) \\ & = √2 \end{align}$$

Since $EK$ is 1, and $∠EKH$ is a right-angle, $KH$ is also 1, and $∠EHK = π/4$.

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:

$$\begin{align} EH^2 & = HK^2 + EK^2 \\ & = (r-1-√3)^2 + 1 \\ & = r^2 - 2r(1-√3) + (1-√3)^2 + 1 \\ & = r^2 - 2r(1-√3) + 5-2√3 \\ EH = & √(r^2 - 2r(1-√3) + 5-2√3) \end{align}$$

$r$ in terms of $h$ (the sagitta, $EL$) and $c$ (the chord length, $DF$) via the intersecting chord theorem:

$$\begin{align} r & = c^2/8h + h/2 = HL \\ c & = DF = 2 \\ h & = EL = HL - EH \\ & = r - √(r^2 - r(2-2√3) + 5-2√3) \\ r & = 4/8h + h/2 = 1/2h + h/2 \\ 2hr & = h^2 + 1 \\ h^2 - 2hr + 1 & = 0 \end{align}$$

Expand out $h^2$ and $2hr$ because then it is very clean to complete the equation above:

$$\begin{align} h^2 & = r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) + (r^2 - r(2-2√3) + 5-2√3) \\ & = 2r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) + 2r(1-√3) + 5-2√3 \\ 2hr & = 2r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) \end{align}$$

Then we have:

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

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:

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

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:

$$\begin{align} & EH^2 = HK^2 + EK^2 = (r-1-√3)^2 + 1 = r^2 - 2r(1-√3) + (1-√3)^2 + 1 \\ & EH^2 = r^2 - 2r(1-√3) + 5-2√3 \\ & EH = √(r^2 - 2r(1-√3) + 5-2√3) \end{align}$$

$r$ in terms of $h$ (the sagitta, $EL$) and $c$ (the chord length, $DF$) via the intersecting chord theorem:

$$\begin{align} r & = c^2/8h + h/2 = HL \\ c & = DF = 2 \\ h & = EL = HL - EH \\ & = r - √(r^2 - r(2-2√3) + 5-2√3) \\ r & = 4/8h + h/2 = 1/2h + h/2 \\ 2hr & = h^2 + 1 \\ h^2 - 2hr + 1 & = 0 \end{align}$$

Expand out $h^2$ and $2hr$ because then it is very clean to complete the equation above:

$$\begin{align} h^2 & = r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) + (r^2 - r(2-2√3) + 5-2√3) \\ h^2 & = 2r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) + 2r(1-√3) + 5-2√3 \\ 2hr & = 2r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) \end{align}$$

Then we have:

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

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:

$$\begin{align} EH^2 & = HK^2 + EK^2 \\ & = (r-1-√3)^2 + 1 \\ & = r^2 - 2r(1-√3) + (1-√3)^2 + 1 \\ & = r^2 - 2r(1-√3) + 5-2√3 \\ EH = & √(r^2 - 2r(1-√3) + 5-2√3) \end{align}$$

$r$ in terms of $h$ (the sagitta, $EL$) and $c$ (the chord length, $DF$) via the intersecting chord theorem:

$$\begin{align} r & = c^2/8h + h/2 = HL \\ c & = DF = 2 \\ h & = EL = HL - EH \\ & = r - √(r^2 - r(2-2√3) + 5-2√3) \\ r & = 4/8h + h/2 = 1/2h + h/2 \\ 2hr & = h^2 + 1 \\ h^2 - 2hr + 1 & = 0 \end{align}$$

Expand out $h^2$ and $2hr$ because then it is very clean to complete the equation above:

$$\begin{align} h^2 & = r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) + (r^2 - r(2-2√3) + 5-2√3) \\ & = 2r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) + 2r(1-√3) + 5-2√3 \\ 2hr & = 2r^2 - 2r√(r^2 - r(2-2√3) + 5-2√3) \end{align}$$

Then we have:

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