First we show that $AE=AF$. Let $BB_1$meet $AM$ at $B'$. Then $AE=AB_1\frac{AB'}{B'B_1}=\frac{AB.cosA}{sin{MAC}}=\frac{AB.AM.cosA}{MC.sinC}$ and similarly $AF=\frac{AC.AM.cosA}{MB.sinB}$ and so $AE=AF$ Let $k_1,k_2$ tangent to $k$ at $X_1,X_2$ and to $EF$ at $Y_1,Y_2$. Let the centers of $k,k_1,k_2$ be $O,O_1$. Then $X_1,O_1,O$ collinear. Since $AE=AF$, hence $M$ is midpoint of arc $EF$ and so the homothety from $X_1$ sending $k_1$ to $k$ sends $Y_1$ to $M$. Therefore $\angle X_1X_2Y_2=\frac{1}{2}\angle X_1OM=\frac{1}{2}\angle X_1O_1Y_1=180-\angle X_1Y_1Y_2\Rightarrow X_1X_2Y_1Y_2$ concyclic $\Rightarrow MY_1.MY_2=MX_1.MX_2\Rightarrow M$ has same power to $k_1,k_2$ and so $M,P,Q$ collinear.

$k_1,k_2$ should touch the arc $EF$ which contains $M$, as opposed to touching the arc which doesn't.

then i don't think they are collinear

To avoid trigo calculations when determining $AE=AF$: take $M, N$ the projections of $B, C$ onto $AM$ respectively; $BM=CM\implies BN=CP\ (\ 1\ )$. Next: $\triangle AC_1F\sim\triangle BNA\implies\frac{AB}{AF}=\frac{BN}{AC_1}$, hence $AF=\frac{AB\cdot AC_1}{BN}\ (\ 2\ )$. Similarly $AE=\frac{AC\cdot AB_1}{CP}\ (\ 3\ )$. But from $\triangle AB_1B\sim\triangle AC_1C$ we get $AB\cdot AC_1=AC\cdot AB_1\ (\ 4\ )$. From $(2)$ and $(3)$, with $(1)$ and $(4)$ we see $AE=AF$ Next, we may use an inversion of pole $M$ and power $PE^2=PF^2$, i.e. this inversion sends the circle $k$ to line $EF$, and preserves the circles $k_1, k_2$; keeping the notations from post #2, we get the same $X_1X_2Y_2Y_1$ cyclic, leading to the fact that $M$ belongs to the radical axis of circles $k_1, k_2$, and we are done. Best regards, sunken rock

Two more proofs of $AE = AF$: 1. Let $P$ be a point so that $AP = AB, AP \perp AB$, and $P,C$ are on opposite sides of $AB$. Similarly, construct $Q$ so that $QA = AC, QA \perp AC$, and $Q,B$ on opposite sides of $AC$. Let $H = BB_1 \cap CC_1$ be the orthocenter of $\triangle ABC$. Now, it is well known that $AM \perp PQ$, and that $AH$ passes through the midpoint of $PQ$. Hence, we have that $\triangle FHE$ and $\triangle PAQ$ are homothetic (having parallel sides). Thus, the $A-$ median of $\triangle PAQ$ must be parallel to the $H -$ median of $\triangle FHE$. But the median of $\triangle PAQ$ passes through $H$, hence $HA$ is the median of $\triangle FHE$, and thus $A$ is the midpoint of $EF$. 2. Let $H$ be the orthocenter of $B_1C_1$ intersect $BC$ at $T$, and let $AH$ intersect $BC$ at $A_1$. The points $B,C_1,B_1,C$ all lie on a circle with center $M$. Thus, the polar of $A = BC_1 \cap CB_1$ must be $TH$, hence $AM \perp TH$. Hence, $TH || EF$. Now, $(B,C;A_1,T) = -1$, thus $(HB,HC;HA_1,HT) = -1$. Now, intersecting these lines with $EF$, we get $(F,E;A, \infty) = -1$, and so $AE = AF$.

Lemma: Circles $\omega_1$ and $\omega_2$ intersect each other at $P$, $Q$ and are internally tangent to circle $\Omega$ at $R,S$ respectively. A common tangent $\ell$ of $\omega_1$ and $\omega _2$ meets $\Omega$ at $E$ and $F$. Then line $PQ$ passes through $M$, the midpoint of arc $EF$ not containing $R,S$.

Let $H$ be the orthocenter of $ABC$, $A_1=AH\cap BC$ and $X=BC\cap B_1C_1$. From Brochard's theorem, $XH\perp AM\implies XH||EF\implies XH\cap EF=P_\infty$. So $-1=(C,B;A_1,X)\stackrel {H}{=} (F,E;A,P_\infty)\implies AE=AF$. Since $MA\perp EF$, $M$ is the midpoint of arc $EMF$ of $\bigcirc MEF$. So from our lemma, it follows $M\in PQ$.

Let $\angle BAM = \alpha_1=\angle AEB, \angle MAC = \alpha_2=\angle AFC$. Now $AE = \frac{AB_1}{sin \alpha_2} = \frac{AB cos \angle A}{sin \alpha_2}$ and in the same way, $AF = \frac{AC cos \angle A}{sin \alpha_1}$. Using sine law in $ABM$ and $AMC$ and comparing, we get $\frac{AB}{sin \alpha_2}=\frac{AC}{sin \alpha_1}$. So, $AE=AF$ and as $MA \perp EF, M$ is the midpoint of the arc $EF$ of $k$ that contains $M$. So if $O$ is the center of $\odot EFM$, then $O, A, M$ collinear. Now, let $k_1$ touches the arc $EF$ and line $EF$ at $P_1$ and $Q_1$ and $k_2$ at $P_2$ and $Q_2$, respectively. Consider the homothety with center $P_1$ that takes $k_1$ to $\odot EFM$ and let $P_1Q_1$ intersects $\odot EFM$ again at $M'$, let $l$ be the line parallel to $EF$ passing through $M'$, now the homothety says that $l$ touches $\odot EFM$ as $EF$ touches $k_1$, so $M'$ is the midpoint of the arc $EF$ that doesn't contain $P_1$. So $M' \equiv M$. Similarly, $P_2, Q_2, M$ collinear. Let $O_2$ is center of $k_2$, so because of the homothety $O_2Q_2 \parallel OM$. Now do some easy angle chasing $\Rightarrow$ $\angle P_2Q_2E = \frac{1}{2}\angle P_2O_2Q_2 = \frac{1}{2}\angle P_2OM = \angle P_2P_1Q_1 \implies P_1P_2Q_1Q_2$ concyclic And thus $MP_1.MQ_1=MP_2.MQ_2 \implies M$ lies on the radical axis $PQ$ of $k_1$ and $k_2$ $\rightarrow$ done!

By easy angle chasing we get $\triangle{AHF} \sim \triangle{MBA}$ and $\triangle{AHE} \sim \triangle{MCA}$ so we get $\frac{AH}{AF}=\frac{BM}{AM}=\frac{CM}{AM}=\frac{AH}{AE} \implies AE=AF \implies ME=MF$. Taking the circles $(M),(F),(k_1),(k_2)$ and applying Casey's theorem we get $MF \cdot XY+t_{k_2} \cdot FY=FX \cdot t_{k_1}$ Again taking circles $(M),(E),(k_2),(k_1)$ and applying Caesy's theorem we get $ME \cdot XY+t_{k_1} \cdot EX=t_{k_2} \cdot EY$ Here $t_{k_1},t_{k_2}$ refer to the tangents drawn from $M$ to $k_1,k_2$ respectively.Subtracting the two equations and using the fact that $ME=MF$ we get $t_{k_1}=t_{k_2}$.So $M$ lies on the radical axis $PQ$ of $k_1$ and $k_2$,as desired.

Proof of $ AE=AF $ : Let $ H $ be the orthocenter of $ \triangle ABC $ . Since $ AB \perp HC_1, AC \perp HB_1, BC \perp HA $ , so from $ (AB, AC; BC, AM)=-1 $ we get $ (HC_1, HB_1; HA, EF)=-1 $ . ie. $ A $ is the midpoint of $ EF $

Let $K_1,K_2$ touch. $K$ respectively at $J,L$ Let $K_1,K_2$ touch segment$EF$ respectively at points $T,Z$ We obtain that $pQ$ passes from $M$ .so it must have the same power between $K_1,K_2$. We know that the only point that has the same power between $K_1,K_2$ and it is on $K$ is the interestion between $JT,LZ$ or we can say that it is the midpoint of arc $FE$ . Let ,$\angle C_1BB_1=\angle B_1CC_1=a$ $\frac{sin\angle a}{AE}=\frac{sin\angle AEB_1}{AB}=\frac{sin\angle MAC}{AB}=\frac{sin\angle BAM}{AC}=\frac{sin\angle AFC}{AC}=\frac{sin\angle a}{AF}$ So $AE=AF$