A circle and secants problem

We note that the circle is the circumcircle of $\triangle BCD$ . Then the circumradius is given by:

$2R = \frac {CD}{\sin \angle CBD}$

By cosine rule , we have:

$\begin{aligned} CD^2 & = 8^2 + 6^2 - 2(8)(16) \cos 60^\circ = 192 \\ \implies CD & = 8 \sqrt 3 \end{aligned}$

And

$\begin{aligned} BC^2 & = 8^2 + 6^2 - 2(8)(6) \cos 60^\circ = 52 \\ \implies BC & = 2 \sqrt {13} \end{aligned}$

By sine rule , we have:

$\begin{aligned} \frac {\sin \angle ABC}{CA} & = \frac {\sin \angle CAB}{BC} \\ \implies \sin \angle ABC & = \frac {\sin 60^\circ}{2\sqrt {13}} \times 8 = \sqrt {\frac {12}{13}} \end{aligned}$

Note that $\sin \angle CBD = \sin (180^\circ - \angle ABC) = \sin \angle ABC = \sqrt{\dfrac {12}{13}}$ and

$\begin{aligned} 2 R & = \frac {CD}{\sin \angle CBD} = \frac {8\sqrt 3}{\sqrt{\frac {12}{13}}} = 4 \sqrt {13} \\ \implies R & = 2 \sqrt {13} \\ R^2 & = \boxed {52} \end{aligned}$

Having forgotten all about power of a point and intersecting secants, I did this with coordinate geometry:

Taking $A$ as the origin, $B$ on the axis and using the fact that $\angle A=60^\circ$ , we can locate points $B,C,D$ : $B(6,0),\;\;C(4,4\sqrt3),\;\;D(16,0)$

These three points lie on the circle, and are enough to define it. If the circle's centre is at $(u,v)$ and its radius is $R$ , we have $(u-6)^2+v^2=R^2,\;\;(u-4)^2+(v-4\sqrt3)^2=R^2,\;\;(u-16)^2+v^2=R^2$

The first and last of these tell us $u=11$ (alternatively, this is just the perpendicular bisector of the chord $BD$ ).

So $25+v^2=R^2\;\;\text{and}\;\;49+(v-4\sqrt3)^2=R^2$

Subtracting, we find $v=3\sqrt3$ , and then substituting into any of the equations gives $R^2=\boxed{52}$ .

From the intersecting secants theorem , we can determine CE

$8(8 + CE) = 6 (16)$

so that $CE = 4$

$\overline{BC}$ can be determined from the law of cosines

$\overline{BC}^2 = 8^2 + 6^2 - 96 \cos A = 64 + 36 - 48 = 52$

$\overline{DE}$ can be determined from the law of cosines

$\overline{DE}^2 = 16^2 + 12^2 - 2(16)(12) \cos A = 256 + 144 - (192) = 208$

The radius of the circle is the circumradius of cyclic quadrilateral $BDEC$ , and this is given by Parameshvara's formula

$R = \dfrac{1}{4} \sqrt{ \dfrac{(ab + cd)(ac + bd)(ad+bc)} {(s - a) (s - b)(s- c)(s-d) } }$

where $a,b,c,d$ are the lengths of the four sides of the cyclic quadrilateral and $s$ is the semi-perimeter , $s = \frac{1}{2} (a + b + c + d)$

So we can set

$a = \overline{BC} = \sqrt{52}$

$b = \overline{BD} = 10$

$c = \overline{DE} = \sqrt{208}$

$d = CE = 4$

Pluggin these into the formula we get

$R = \sqrt{52}$

Thus the answer is $R^2 = \boxed{52}$

By power of a point, $AB \cdot AD = AC \cdot AE$ , so $AE = \frac{6 \cdot 16}{8} = 12.$

Since $\angle EAB = 60^\circ$ and $AE = 2AB$ , triangle $EAB$ is a $30^\circ-60^\circ-90^\circ$ triangle, so $EB = 6 \sqrt{3}$ .

Since $\angle EBD = 90^\circ$ , by Pythagoras, $ED^2 = (6 \sqrt{3})^2 + 10^2 = 208$ . The fact that $\angle EBD = 90^\circ$ also implies $ED$ is a diameter of the circle, so $R^2 = \frac{ED^2}{4} = 52.$

Let the center of the circle be $O$ , and let the midpoints of $BD$ and $CE$ be $F$ and $G$ , respectively. Then $R = OB = OC = OD = OE$ .

By the intersecting secant theorem , $AC \cdot AE = AB \cdot AD$ , or $8(8 + CE) = 6 \cdot 16$ , which solves to $CE = 4$ .

Since $F$ is the midpoint of $BD$ , and $BD = 10$ , $BF = 5$ , and by the Pythagorean Theorem on $\triangle BFO$ , $OF = \sqrt{R^2 - 25}$ .

Similarly, since $G$ is the midpoint of $CE$ , and $CE = 4$ , $CG = 2$ , and by the Pythagorean Theorem on $\triangle CGO$ , $OG = \sqrt{R^2 - 4}$ .

From $\triangle FAO$ , $\tan FAO = \frac{OF}{AF} = \frac{\sqrt{R^2 - 25}}{11}$ , and from $\triangle GAO$ , $\tan GAO = \frac{OG}{AG} = \frac{\sqrt{R^2 - 4}}{10}$ .

By the tangent addition formula , $\frac{\tan FAO + \tan GAO}{1 - \tan FAO \cdot \tan GAO} = \tan(FAO + GAO)$ or $\frac{\frac{\sqrt{R^2 - 25}}{11} + \frac{\sqrt{R^2 - 4}}{10}}{1 - \frac{\sqrt{R^2 - 25}}{11} \cdot \frac{\sqrt{R^2 - 4}}{10}} = \tan 60° = \sqrt{3}$ , which solves to $R^2 = \boxed{52}$ .