Two Chords and an Angle

the first thing to do is to find DB. then using a combination of law of sine and cosine find $\sin\theta$ , and end it by using $2 r \sin\theta = DB$ .notice inscribed angle theorem was used. first $DB^2 = 14^2+196^2-2\dfrac{8}{17} *14*196 \to DB= 350 \sqrt{\dfrac{5}{17}}$ then $BC^2=56^2+196^2+2\dfrac{8}{17} *56*196 \to BC = 420 \sqrt{\dfrac{5}{17}}$ and by laws of sine $\dfrac{196}{\sin\theta} = \dfrac{420 \sqrt{\dfrac{5}{17}}}{\sin(\arccos(-8/17))} \to \dfrac{1}{\sin\theta} = \dfrac{17}{7} \sqrt{\dfrac{5}{17}}$ this gives $r = \dfrac{DB}{2\sin\theta}=\dfrac{1}{2} \times 350 \sqrt{\dfrac{5}{17}}\times \dfrac{17}{7} \sqrt{\dfrac{5}{17}} = \boxed{125}$

Let $FG$ be the perpendicular bisector of $CD$ at $G$ , let $HI$ be the perpendicular bisector of $AB$ at $I$ , and let $O$ be the intersection of $FG$ and $HI$ , and therefore the center of the circle.

By the intersecting chords theorem , $AE \cdot EB = CE \cdot ED$ , so $4 \cdot 196 = CE \cdot 14$ , which means $CE = 56$ . Since $I$ is a bisector of $AB$ , $IB = 100$ and $EI = 96$ . Since $G$ is a bisector of $CD$ , $CG = 35$ and $GE = 21$ .

Let $E$ be the origin and $EB$ be the positive $x$ -axis. Since $\angle CEB = \cos^{-1}(-\frac{8}{17})$ , and $\angle AEG$ is supplementary to it, $\angle AEG = \cos^{-1}(\frac{8}{17})$ . Since $EG = 21$ , it has an $x$ -component of $21\cos(\cos^{-1}(\frac{8}{17})) = \frac{168}{17}$ , and a $y$ -component of $\sqrt{21^2 - (\frac{168}{17})^2} = \frac{315}{17}$ , and so $G$ has coordinates $(-\frac{168}{17}, -\frac{315}{17})$ .

Chord $CD$ has a slope of $\tan(\cos^{-1}(\frac{8}{17})) = \frac{\sqrt{17^2-8^2}}{8} = \frac{15}{8}$ , and since $FG$ is perpendicular to $CD$ , $FG$ has a slope of $-\frac{8}{15}$ , and therefore an equation of $y + \frac{315}{17} = -\frac{8}{15}(x + \frac{168}{17})$ .

Since $HI$ is perpendicular to the $x$ -axis, and $EI = 96$ , $HI$ is a vertical line with an equation of $x = 96$ .

Therefore, since $O$ is the intersection of $FG$ and $HI$ , its coordinates are the solution to $y + \frac{315}{17} = -\frac{8}{15}(x + \frac{168}{17})$ and $x = 96$ , which is $(96, -75)$ , which means $IO = 75$ .

Finally, by Pythagorean's Theorem on $\triangle OHB$ , the radius $OB = \sqrt{IO^2 + IB^2} = \sqrt{75^2 + 100^2} = \boxed{125}$ .

Let $O$ be the center of the circle, $M$ , $N$ the midpoints of $AB$ and $CD$ respectively. Then $OM\bot AB$ and $ON\bot CD$ , so $\angle ONE+\angle OME =\pi$ . Hence $OMEN$ is a cyclic quadrilateral. Consequently, $\angle MEN+\angle MON=\pi \Rightarrow \cos \left(MON \right)=-\cos \left( MEN \right)\Rightarrow \cos \left( MON \right)=\dfrac{8}{17}$

By the intersecting chords theorem , same way as @David Vreken did, we have $AE \cdot EB = CE \cdot ED$ , so $4 \cdot 196 = CE \cdot 14$ , thus, $CE = 56$ .
Since $M$ and $N$ are midpoints, we get $EM=96$ , $MB=100$ , $CN=35$ , $NE=21$ .

Let $\angle MOE={{\omega }_{1}}$ , $\angle NOE={{\omega }_{2}}$ , $OB=R$ and $OE=d$ . Then, $\sin {{\omega }_{1}}=\dfrac{96}{d}$ , $\sin {{\omega }_{2}}=\dfrac{21}{d}$ , $\cos {{\omega }_{1}}=\sqrt{1-{{\left( \dfrac{96}{d} \right)}^{2}}}$ and $\cos {{\omega }_{2}}=\sqrt{1-{{\left( \dfrac{21}{d} \right)}^{2}}}$ .

Now, we have an equation for $d$ :

$\begin{aligned} \cos \left( \angle MON \right)=\cos \left( {{\omega }_{1}}+{{\omega }_{2}} \right) & \Leftrightarrow \dfrac{8}{17}=\cos {{\omega }_{1}}\cdot \cos {{\omega }_{2}}-\sin {{\omega }_{1}}\cdot \sin {{\omega }_{2}} \\ & \Leftrightarrow \dfrac{8}{17}=\sqrt{1-{{\left( \dfrac{96}{d} \right)}^{2}}}\cdot \sqrt{1-{{\left( \dfrac{21}{d} \right)}^{2}}}-\dfrac{96}{d}\cdot \dfrac{21}{d} \\ & \Leftrightarrow \cdots \\ & \Leftrightarrow {{d}^{2}}=14841 \\ \end{aligned}$

Finally comes Pythagorean Theorem:

$\begin{aligned} O{{B}^{2}}=M{{B}^{2}}+O{{M}^{2}} & \Rightarrow {{R}^{2}}={{100}^{2}}+\left( {{d}^{2}}-{{EM}^{2}} \right) \\ & \Rightarrow {{R}^{2}}={{100}^{2}}+\left( 14841-{{96}^{2}} \right) \\ & \Rightarrow R=\boxed{125} \\ \end{aligned}$