Distance from the Center

Observe the following: Angle CFE = 90 deg - Angle DBF = 90 deg - Angle OBE = 90 deg - Angle OEB (because OE = EB, radii of circle) = Angle CEF. because Angle CEO is 90 deg (tangent to circle) Angle CFE = Angle CEF implies that Triangle CEF is isoceles, with CE = CF = 110 Finally, by Pythagoras' theorem on Triangle CEO, $OC = \sqrt{ OE^2 + EC^2} =\sqrt {96^2 + 110^2} = 146$ (shown)

Consider $\angle DBF=x ^ \circ$ We note that $\angle BDF=90 ^ \circ$ . So that means $\angle BFD=(90-x) ^ \circ$ . So $\angle EFC=(90-x) ^ \circ$ .

Now E is tangent to the circle so, $\angle OEC=90 ^ \circ$ . Furthermore, $\angle OEB=\angle OEF=\angle DBF=x ^ \circ$ as triangle $OEB$ is an isosceles triangle. So $\angle FEC=(90-x) ^ \circ=\angle EFC$ . Hence triangle $ECF$ is an isosceles triangle, with $FC=EC=110$

Now $OE=192/2=96$ and with triangle $OEC$ being a right angle triangle, by pythagoras theorem, $OC^2=110^2+96^2$ Hence $OC=146$

Let angle ABE = \theta. It can be seen that angle DFB = 180^ \circ - angle DBF - angle BDF = 180 ^ \circ - \theta - 90 ^ \circ = 90 ^ \circ- \theta. Moreover, angle EFC = angle DFB = 90 ^ \circ -\theta. (vertically opposite angles)

Now, angle OEB = angle OBE = \theta. (OE = OB = radius of circle). Since OE (radius of circle) is perpendicular to CE (tangent of circle), angle OEC = 90^ \circ and angle CEF = angle CEO - angle OEB = 90 ^ \circ -\theta. Since angles EFC and FEC are equal, then CE = CF = 110. Therefore, by Pythagoras' Theorem, CO = \sqrt{OE^2 + CE^2} = \sqrt{96^2+ 110^2} = \sqrt{21316} = 146. (ans)

Let angle FEC be x Since EC is tangent, angle OEC = 90, Thus angle OEB = OCE-FEC = 90-x Since OE=OB, angle OBE = angle OEB = 90-x Since angle CDB = 90, angle DFB = 180- CDB - OBE = x angle EFC = angle DFB = x thus angle FEC = angle EFC and CE = CF = 110 OC = (OE^2+EC^2)^(0.5) by Pythagoras Theorem since OE = 192/2 = 96 and EC = 110, OC = 146

Continuing from the problem description, extend $CD$ so that it intersects $\Gamma$ at $X$ between $BE$ and at $Y$ on the other side of $AB$ . Let $ABE=\theta$ . Then, $BFD=CFE=90-\theta$ . Now, we will consider arcs. We know $AE=2\theta$ . Note that $AY=AX=AE+EX$ , and $ECF=\frac{1}{2}(EY-EX)=\frac{1}{2}(AE+AY-EX)=\frac{1}{2}(AE+AE)=AE=2\theta$ . We see that $ECF$ is an isosceles triangle, so $CE=CF=110$ . Drawing radius $OE$ , we know that $OEC$ is a right triangle because $CE$ is a tangent. Pythagorean theorem gives $OC=146$ .

Because $AB$ is a diameter, $\angle AEB = 90^\circ$. $CE$ is tangent to $\Gamma$, so $\angle OEC = 90^\circ$. We then have $\angle AEB = \angle AEO + \angle OEB = 90^\circ = \angle OEC = \angle FEC + \angle OEB$. Therefore, $\angle AEO = \angle FEC$. $\triangle AOE$ is isosceles with $AO = OE$, so $\angle AEO = \angle EAO$. $\triangle BDF$ is similar to $\triangle BEA$ by $AA$ similarity since $\angle BDF = 90^\circ = \angle BEA$ and $\angle FBD = \angle ABE$. Therefore, $\angle EAO = \angle EAB = \angle DFB$. $\angle DFB = \angle EFC$ since they are vertical angles. We conclude that $\angle FEC = \angle EFC$, and so $\triangle ECF$ is isosceles with $CE = CF = 110$. By Power of Point, the power of $C$ with respect to circle $\Gamma$ is both $110^2$ and $CO^2 - R^2$, where $R$ is the radius of $\Gamma$. Therefore, $110^2 = CO^2 - R^2 = CO^2 - (192/2)^2 = CO^2 - 96^2$. Solving for $CO$, we obtain $CO = \sqrt{110^2 + 96^2} = 146$.

AB is diameter , E on the circumference. $\angle BEA=90^\circ$ CE touches the circle at E point , $\angle OEC=90^\circ$ now , $\angle BEA= \angle OEC$ $\Rightarrow \angle BEO + \angle OEA=\angle BEO + \angle BEC$ $\Rightarrow \angle OEA= \angle BEC$ $\Rightarrow \angle OAE= \angle BEC [\because OE=OA \therefore \angle OEA=\angle OAE]$ $\Rightarrow \angle BAE= \angle FEC$ in $\Delta BFD$ and $\Delta BAE$ , $\angle FBD=\angle EBA$ and $\angle FDB=\angle AEB(=90^\circ)$ $\therefore \angle BFD=\angle BAE$ $\angle BAE= \angle FEC$ $\Rightarrow \angle BFD= \angle FEC$ $\Rightarrow \angle CFE= \angle FEC$ $\therefore CF=CE ; CE=110$ now , assume that OC intersects the circle at X and extending at Y . Using Power of a point theorem , $CX.CY=CE^2$ $\Rightarrow (OC-OX)(OC+OY)=CE^2$ diameter AB=192 , so radius , $OA=OB=OX=OY=\frac{192}{2}=96$ $(OC-OX)(OC+OY)=CE^2$ $\Rightarrow (OC-96)(OC+96)=CE^2$ $\Rightarrow OC^2-96^2=110^2$ $\Rightarrow OC^2=110^2+96^2=146^2$ $\therefore OC=146$

CFE=DFB=BAE=CEF

$\begin{array}{lll} \angle CEB & = \angle EAB &\qquad & \mbox{by alternate segment theorem}\\ & = 90^\circ - \angle EBA & & \mbox{since }\angle AEB=90^\circ \\ & = \angle BFD & & \mbox{since }\angle FDB=90^\circ \\ & = \angle EFC & & \mbox{opposite angles}\\ \end{array}$

Hence, $CEF$ is an isosceles triangle, so $CE= CF=110$ .

Since, $CE$ is tangent to $\Gamma$ at $E$ , thus $\angle CEO = 90^\circ$ . Therefore, $OC^2 = OE^2 + CE^2 = 96 ^2 + 110 ^2 = 146 ^2 \Rightarrow OC = 146$ .