A hexagon is inscribed in a circle of radius r. Two of its sides have length 1, two have length 2 and last two sides have length 3. Then radius r satisfies the equation

If side of length i subtends angle ${\alpha }_i$, at the
centre, then $2{\alpha }_1+2{\alpha }_2+2{\alpha }_3={360}^{\circ }$
 

Also $\sin \left(\frac{{\alpha }_1}{2}\right)=\frac{1}{2r},\sin \left(\frac{{\alpha }_2}{2}\right)=\frac{1}{r},\sin \left(\frac{{\alpha }_3}{2}\right)=\frac{3}{2r}$
$\begin{array}{l}\cos \frac{{\alpha }_1}{2}=\frac{\sqrt{4r^2-1}}{2r},\cos \left(\frac{{\alpha }_3}{2}\right)=\frac{\sqrt{r^2-1}}{r}\\ \frac{{\alpha }_1}{2}+\frac{{\alpha }_2}{2}={90}^{\circ }-\frac{{\alpha }_3}{2}\\ \Rightarrow \cos \frac{{\alpha }_1}{2}\cos \frac{{\alpha }_2}{2}-\sin \frac{{\alpha }_1}{2}\sin \frac{{\alpha }_2}{2}=\sin \frac{{\alpha }_3}{2}\\ \Rightarrow \frac{\sqrt{4r^2-1}\sqrt{r^2-1}}{2r^2}-\frac{1}{2r^2}=\frac{3}{2r}\\ \Rightarrow \sqrt{\left(4r^2-1\right)\left(r^2-1\right)}=3r+1\\ \Rightarrow 4r^4-5r^2+1=9r^2+6r+1\\ \Rightarrow 2r^3-7r-3=0\end{array}$