Cyclic quadrilateral and the diameter

Let the diameter of the circle be $AD=d$ . Since $AD$ is a diameter, $\angle ABD = \angle ACD = 90^\circ$ . Also since $ABCD$ is a cyclic quadrilateral , we can apply Ptolemy's theorem .

$\begin{aligned} AC \times BD & = BC\times AD + AB \times CD \\ \sqrt{d^2-6^2}\times \sqrt{d^2-14^2} & = 6d + 14 \times 6 & \small \blue{\text{Squaring both sides}} \\ d^4-232d^2+7056 & = 36d^2 + 1008d + 7056 \\ d^3 -268d-1008 & = 0 \\ (d-18)(d+4)(d+14) & = 0 & \small \blue{\text{Since }d > 0} \\ \implies d & = 18 \end{aligned}$

Therefore the radius of the circle is $\dfrac d2 = \dfrac {18}2 = \boxed 9$ .

Let the center of the circle be ' $O$ '. Join $\overline {OB}, \overline {OC}$ . Let $\angle {ODC}=\angle {OCD}=\angle {OCB}=\angle {OBC}=α$ . Then $\angle {OAB}=\angle {OBA}=π-2α$ (Sum of opposite angles of a cyclic quadrilateral is $π$ ). Now, using sine rule we can write $\dfrac{r}{\sin α}=\dfrac{6}{\sin 2α}\implies r=\dfrac{3}{\cos α}$ , where $r$ is the radius of the circle. And $\dfrac{r}{\sin (π-2α)}=\dfrac{14}{\sin (4α-π)}\implies r=-\dfrac{7}{\cos 2α}=-\dfrac{7}{2\cos^2 α-1}$ . Therefore $\dfrac{3}{\cos α}=\dfrac{7}{1-2\cos^2 α}\implies 6\cos^2 α+7\cos α-3=0\implies \cos α=\dfrac{1}{3}$ . Hence $r=\dfrac{3}{\dfrac{1}{3}}=\boxed 9$ .