$E B=\frac{7}{2} \quad O$ is the circumcenter of $\triangle ABC \Longrightarrow AO=CO \Longrightarrow \angle OCA=\angle OAC$. Because $AC$ is an inscribed arc of circumcircle $\triangle AOC, \angle OCA=\angle OFA$. Furthermore $BC$ is tangent to circumcircle $\triangle AOC$, so $\angle OAC=\angle OCB$. However, again using the fact that $O$ is the circumcenter of $\triangle ABC, \angle OCB=\angle OBC$. We now have that $CO$ bisects $\angle ACB$, so it follows that triangle $CA=CB$. Also, by AA similarity we have $EOB \sim EBF$. Thus, $EB^{2}=EO \cdot EF=EC^{2}$ by the similarity and power of a point, so $EB=BC / 2=AC / 2=7 / 2$.