hexagon in a circle

Draw sides of lengths $2$ , $7$ , and $11$ next to each other. Then we have cyclic quadrilateral $ABCD$ with $AD=2r$ being the diameter of the circle. By Ptolemy's theorem ,

$\begin{aligned} AC\cdot BD & = BC \cdot AD + AB \cdot CD \\ \sqrt{(2r)^2-11^2} \cdot \sqrt{(2r)^2-2^2} & = 14r + 22 & \small \blue{\text{Squaring both sides}} \\ (4r^2-121)(4r^2-4) & = 14^2r^2 + 2(14)(22)r + 22^2 \\ 16r^4 - 500 r^2 + 484 & = 196 r^2 + 616 r + 484 \\ 16r^4-696r^2-616r & = 0 & \small \blue{\text{Divide both sides by }8r } \\ 2r^3 - 87r - 77 & = 0 \\ (r-7)(2r^2+14r+11) & = 0 \end{aligned}$

$\implies r = \begin{cases} 7 \\ \frac {3\sqrt 3-7}2 < 0 \\ \frac {-3\sqrt 3-7}2 < 0 \end{cases}$ . Therefore $r=\boxed 7$ .

Let the side of length $2\,,\,7$ and $11$ subtend an angle of $\alpha\,,\,\beta$ and $\gamma$ respectively at the center of circle. Radius of circle is $r$ . Then we get,

$\displaystyle 2r\sin\Big(\frac{\alpha}{2}\Big) = 2\hspace{20pt}\cdots\text{Eq. 1}$

$\displaystyle 2r\sin\Big(\frac{\beta}{2}\Big) = 7\hspace{20pt}\cdots\text{Eq. 2}$

$\displaystyle 2r\sin\Big(\frac{\gamma}{2}\Big) = 11\hspace{20pt}\cdots\text{Eq. 3}$

$\displaystyle 2\alpha + 2\beta + 2\gamma = 360\degree$

$\displaystyle\Rightarrow \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90\degree\hspace{20pt}\cdots\text{Eq. 4}$

Substituting $\gamma$ from Eq. 4 into Eq. 3 we get,

$\displaystyle 2r\sin\Big(90\degree - \frac{\alpha + \beta}{2}\Big) = 11$

$\displaystyle 2r\cos\Big(\frac{\alpha + \beta}{2}\Big) = 11 \hspace{20pt}\cdots\text{Eq. 5}$

Eliminating $r$ from Eq. 1 and Eq. 2 we get,

$\displaystyle \frac{\sin\Big(\frac{\alpha}{2}\Big)}{2} = \frac{\sin\Big(\frac{\beta}{2}\Big)}{7}$

$\displaystyle\Rightarrow 7\sin\Big(\frac{\alpha}{2}\Big) = 2\sin\Big(\frac{\beta}{2}\Big)$

$\displaystyle\Rightarrow 49\sin^2\Big(\frac{\alpha}{2}\Big) = 4\sin^2\Big(\frac{\beta}{2}\Big)\hspace{20pt}\cdots\text{Eq. 6}$

Eliminating $r$ from Eq. 1 and Eq. 5 we get,

$\displaystyle \frac{\sin\Big(\frac{\alpha}{2}\Big)}{2} = \frac{\cos\Big(\frac{\alpha + \beta}{2}\Big)}{11}$

$\Rightarrow 11\sin\Big(\frac{\alpha}{2}\Big) = 2\cos\Big(\frac{\alpha + \beta}{2}\Big)$

$\Rightarrow 11\sin\Big(\frac{\alpha}{2}\Big) = 2\cos\Big(\frac{\alpha}{2}\Big)\cos\Big(\frac{\beta}{2}\Big) - 2\sin\Big(\frac{\alpha}{2}\Big)\sin\Big(\frac{\beta}{2}\Big)$

$\Rightarrow \sin\Big(\frac{\alpha}{2}\Big)\bigg(11 + 2\sin\Big(\frac{\beta}{2}\Big)\bigg) = 2\cos\Big(\frac{\alpha}{2}\Big)\cos\Big(\frac{\beta}{2}\Big)$

Squaring the equation we get,

$\displaystyle\Rightarrow \sin^2\Big(\frac{\alpha}{2}\Big)\bigg(11 + 2\sin\Big(\frac{\beta}{2}\Big)\bigg)^2 = 4\cos^2\Big(\frac{\alpha}{2}\Big)\cos^2\Big(\frac{\beta}{2}\Big)$

Multiplying $49$ on both sides

$\displaystyle 49\sin^2\Big(\frac{\alpha}{2}\Big)\bigg(11 + 2\sin\Big(\frac{\beta}{2}\Big)\bigg)^2 = 4\cos^2\Big(\frac{\beta}{2}\Big)\bigg(49 - 49\sin^2\Big(\frac{\alpha}{2}\Big)\bigg)$

Using Eq. 6 in the above equation we get,

$\displaystyle 4\sin^2\Big(\frac{\beta}{2}\Big)\bigg(11 + 2\sin\Big(\frac{\beta}{2}\Big)\bigg)^2 = 4\cos^2\Big(\frac{\beta}{2}\Big)\bigg(49 - 4\sin^2\Big(\frac{\beta}{2}\Big)\bigg)$

Let $\sin\Big(\frac{\beta}{2}\Big) = t$ . Then

$\displaystyle t^2(11 + 2t)^2 = (1 - t^2)(49 - 4t^2)$

$\displaystyle\Rightarrow 44t^3 + 174t^2 - 49 = 0$

$\displaystyle\Rightarrow (2t - 1)(22t^2 +98t + 49) = 0$

Since $\frac{\beta}{2} \leq 90\degree \Rightarrow \sin\Big(\frac{\beta}{2}\Big) = t \geq 0$

For $t \geq 0$ , second factor can never be $0$ . So

$2t - 1 = 0$

$\displaystyle\Rightarrow t = \sin\Big(\frac{\beta}{2}\Big) = \frac{1}{2}$

Putting this in Eq. 2 we get

$\displaystyle 2r\frac{1}{2} = 7$

$\displaystyle\Rightarrow r =7$

Let the $2$ units chord subtends an angle $α$ at the centre of the circle, and the $7$ units chord subtends an angle $β$ . Then

$\sin (\frac{α}{2})=\frac{1}{r},\sin (\frac{β}{2})=\frac{7}{2r},\cos (\frac{α+β}{2})=\frac{11}{2r}$ .

Solving we get $r=\boxed 7$ units.