Moreover the circumcircle of $ \triangle OBF$ also passes through such a common point. Consider the inversion through pole $ O$ and power $ \overline{OE} \cdot \overline{OF}.$ Circle with diameter $ EF$ is double and the opposite rays of $ OB,OA$ cut $ k'$ and $ k,$ respectively at the inverse images $ A',B'$ of $ A,B.$ Then $ \odot(OBF) \mapsto EB',$ $ \odot(OEA) \mapsto FA'$ and circle with diameter $ AB$ is transformed into the circle with diameter $ A'B'.$ Let $ P \equiv FA' \cap EB'.$ Since $ EB' \perp FA'$ $ \Longrightarrow$ $ P$ is common point of $ FA',EB'$ and the circles with diameters $ A'B'$ and $ EF.$ Hence, $ \odot(OEA),\odot(OBF)$ and circles with diameters $ AB,EF$ concur at the inverse $ P'$ of $ P.$

Ma solution is the following: Denote $ k_1$, $ k_2$ the circles with diametres $ AB$, $ DE$, respectively. Moreover, $ k_1\cap k_2 = \{M,N\}$, where $ M$ is in the inner of angle $ \angle{AOE}$, and $ N_1 = AE\cap k_1$ and $ N_2 = AE \cap k_2$. As $ \angle{AN_1B} = \angle{FN_2E} = 90^{\circ}$ due to Thales' theorem, and $ \triangle{OBF}$ is isosceles with $ \overline{OB} = \overline{OF}$ or $ \angle{FBO} = \angle{OFB}$, we must have $ N = N_1 = N_2$. Thus, we have \[ \angle{FEM} = \angle{FNM} = 90^{\circ} - \angle{MNA} = 90^{\circ} - \angle{MBA} = \angle{BAM} = \angle{OAM}\mbox{,}\] implying that $ AOEM$ is cyclic.

I have made it too much complicated Let $m$ denote the circle with radius $AB$ and $n$ denote the circle with radius $EF$.Draw tangents to $m$ at $A$ and $B$. The tangent at $E$ meets the tangent at $A$ at $M$ and The tangent at $F$ meets the tangent at $B$ at $N$.The points $M$ and $N$ have equal powers w.r.t $m$ and $n$.So $MN$ is the radical axis of $m$ and $n$.Our required point must lie on $MN$.Observe that $M$ lies on the circumcircle of $\bigtriangleup OAE$.But this is not the required point,since its power w.r.t $n$ is $MF^2>0$.Let $P$ be the projection of $O$ on $MN$.This $P$ is the second intersection of the circle $OAE$ with $MN$.We claim this one is the required point. $\angle OPE=\angle EAO$ since $A,O,E,P,M$ are concyclic.and since $OA=OE$ we have $\angle OEA=\angle OAE=\angle OPE$. Also,$\angle FPN=\angle FON=\angle NOB$ since $PONF$ is cyclic and $FN=NB$.Thus $\angle EOB=2\angle OPE=2\angle FPN$.So $\angle FPE=\angle EPN+\angle FPN=\angle EPN+\angle OPE=\frac{\pi}{2}$.So $P\in n$ and since $MN$ is the radical axis,P is also on $m$. QED