[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.64, xmax = 19.02, ymin = -10.32, ymax = 5.14; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((5.017079622156162,-0.0702900807401852),0.6,42.858436552730765,96.22504243845378)--(5.017079622156162,-0.0702900807401852)--cycle, linewidth(1) + qqwuqq); draw(arc((9.287124516792435,-2.6616991692150336),0.6,129.03759007102846,182.40419595675158)--(9.287124516792435,-2.6616991692150336)--cycle, linewidth(1) + qqwuqq); draw(arc((6.977847723380699,3.270642374059604),0.6,-173.77495756154622,-120.4083516758231)--(6.977847723380699,3.270642374059604)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw(circle((6.22,-0.44), 3.7872417403698964), linewidth(1)); draw(circle((5.462152276619301,-4.150642374059604), 4.104554117726785), linewidth(1)); draw((5.017079622156162,-0.0702900807401852)--(8.83601638261342,3.4733258675802405), linewidth(1)); draw((3.3499012003832087,-2.9109781222095554)--(9.287124516792435,-2.6616991692150336), linewidth(1)); draw((3.7807714588156975,1.223303560456394)--(9.51817699151674,-4.779949943924331), linewidth(1)); draw((4.68,3.02)--(2.5273396127849685,-1.2811059770878679), linewidth(1)); draw((4.68,3.02)--(9.287124516792435,-2.6616991692150336), linewidth(1)); draw((-0.4704346677023634,-0.6688517675314166)--(4.68,3.02), linewidth(1)); draw((-0.4704346677023634,-0.6688517675314166)--(5.512055018726986,1.9938748559082409), linewidth(1)); draw((-0.4704346677023634,-0.6688517675314166)--(5.017079622156162,-0.0702900807401852), linewidth(1)); draw((-0.4704346677023634,-0.6688517675314166)--(9.287124516792435,-2.6616991692150336), linewidth(1)); draw((2.5273396127849685,-1.2811059770878679)--(9.51817699151674,-4.779949943924331), linewidth(1)); draw((5.512055018726986,1.9938748559082409)--(3.2149945745387947,-7.585412638186182), linewidth(1)); draw((3.3499012003832087,-2.9109781222095554)--(6.977847723380699,3.270642374059604), linewidth(1)); draw((5.462152276619301,-4.150642374059604)--(-0.4704346677023634,-0.6688517675314166), linewidth(1)); draw((4.68,3.02)--(8.83601638261342,3.4733258675802405), linewidth(1)); draw((4.68,3.02)--(5.462152276619301,-4.150642374059604), linewidth(1)); draw((3.2149945745387947,-7.585412638186182)--(9.287124516792435,-2.6616991692150336), linewidth(1)); draw((5.512055018726986,1.9938748559082409)--(8.83601638261342,3.4733258675802405), linewidth(1)); /* dots and labels */ dot((4.68,3.02),dotstyle); label("$A$", (4.54,3.36), NE * labelscalefactor); dot((2.5273396127849685,-1.2811059770878679),dotstyle); label("$B$", (1.96,-1.6), NE * labelscalefactor); dot((9.287124516792435,-2.6616991692150336),dotstyle); label("$C$", (9.5,-2.78), NE * labelscalefactor); dot((5.462152276619301,-4.150642374059604),linewidth(4pt) + dotstyle); label("$M$", (5.66,-3.98), NE * labelscalefactor); dot((5.017079622156162,-0.0702900807401852),linewidth(4pt) + dotstyle); label("$I$", (5.38,-0.18), NE * labelscalefactor); dot((-0.4704346677023634,-0.6688517675314166),linewidth(4pt) + dotstyle); label("$Q$", (-1.04,-0.8), NE * labelscalefactor); dot((3.4398904867702496,2.1318069706821787),linewidth(4pt) + dotstyle); label("$G$", (3.1,2.42), NE * labelscalefactor); dot((6.977847723380699,3.270642374059604),linewidth(4pt) + dotstyle); label("$N$", (7,3.54), NE * labelscalefactor); dot((8.83601638261342,3.4733258675802405),linewidth(4pt) + dotstyle); label("$F$", (8.92,3.64), NE * labelscalefactor); dot((3.3499012003832087,-2.9109781222095554),linewidth(4pt) + dotstyle); label("$K$", (3.08,-3.36), NE * labelscalefactor); dot((3.7807714588156975,1.223303560456394),linewidth(4pt) + dotstyle); label("$X$", (3.46,1.4), NE * labelscalefactor); dot((5.512055018726986,1.9938748559082409),linewidth(4pt) + dotstyle); label("$Y$", (5.6,2.16), NE * labelscalefactor); dot((9.51817699151674,-4.779949943924331),linewidth(4pt) + dotstyle); label("$B_1$", (9.8,-4.96), NE * labelscalefactor); dot((3.2149945745387947,-7.585412638186182),linewidth(4pt) + dotstyle); label("$C_1$", (3.74,-7.6), NE * labelscalefactor); dot((7.76,-3.9),linewidth(4pt) + dotstyle); label("$D$", (7.7,-4.44), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $D$ be the antipode of $A$ in $\odot(ABC)$, $N$ be the antipode of $M$ in $\odot(ABC)$, $K$ be the touchpoint of $A$-mixtilinear incircle of $\bigtriangleup ABC$ with $\odot(ABC)$, and $G$ be the point on $\odot(ABC)$ such that $\angle AGI = 90^{\circ}$ (i.e. the so-called Sharky-Devil point). Let $QF$ cut $AB, AC$ at $X, Y$ respectively. Claim 01. $Q, G, N, F$ are concyclic. Proof. It's known that $A, G, Q$ are collinear (e.g. by considering the radical center of three circles $\odot(AI)$, $\odot(ABC)$, and $\odot(BIC)$). Since $\overline{IG} \perp \overline{AQ}$, we obtain $AG \times AQ = AI^2$. A property of mixtilinear touchpoint tells us that $K, I, N$ are collinear, and $\overline{AK}, \overline{AE}$ are isogonal with respect to $\angle BAC$. Therefore, $\angle ANK = \angle ACK = \angle AEB = \angle AIF \implies \angle ANI = \angle AIF$. Therefore, $\bigtriangleup AIN \sim \bigtriangleup AFI$, i.e. $AI^2 = AN \times AF$. Therefore, $AG \times AQ = AN \times AF \implies Q, G, N, F$ are concyclic. $\square$ Claim 02. $\overline{XI}$ is the external angle bisector of $\angle AXY$, and similarly $\overline{YI}$ is the external angle bisector of $\angle AYX$. Proof. The inversion with center $A$ and radius $AI$ swaps $\{G, Q\}$ and $\{N, F\}$. Therefore, $\odot(AGN) \equiv \odot(ABC)$ maps to $\overline{QF}$ under this inversion, i.e. it swaps $\{X, B\}$ and $\{Y, C\}$. So, $AX \times AB = AI^2 \implies \bigtriangleup AXI \sim \bigtriangleup AIB$. We also know that $\angle AXY = \angle ACB$, since $AX \times AB = AY \times AC$. By angle chasing, we obtain $\angle BXI = 180^{\circ} - \angle AXI = 180^{\circ} - \angle AIB = 180^{\circ} - (90^{\circ} + \frac{\angle ACB}{2}) = 90^{\circ} - \frac{\angle ACB}{2}$. Therefore, $\overline{XI}$ is the external angle bisector of $\angle AXY$. Similarly for $\overline{YI}$. $\square$ Now, we will locate $B_1$, the center of $\omega_b$, and $C_1$, defined similarly. Since $\omega_b$ is tangent to $\overline{AB}$ at $B$, we must have $\overline{B_1B} \perp \overline{AB}$. Moreover, since $\omega_b$ is also tangent to $\overline{FQ}$, it must lie on the bisector of the angle formed by $\overline{FQ}$ and $\overline{AB}$, and since it intersects the interior of $\bigtriangleup ABC$, $B_1 \in \overline{XI}$. This means $\angle IB_1B = \angle XB_1B = 180^{\circ} - 90^{\circ} - (90^{\circ} - \frac{\angle ACB}{2}) = \frac{\angle ACB}{2} = \angle ICB \implies B_1 \in \odot(BIC)$. Similarly, $C_1 \in \odot(BIC)$. Since $\angle B_1MC_1 = 2 \times \angle B_1IC_1 = 2 \times \angle XIY = 2 \times (90^{\circ} - \frac{\angle XAY}{2}) = 180^{\circ} - \angle BAC = \angle BMC$, so $B_1C_1 = BC$, i.e. $BB_1 = CC_1$. Therefore, $$ \text{Pow}(M, \omega_b) = B_1M^2 - R_{\omega_b}^2 = B_1M^2 - BB_1^2 = C_1M^2 - CC_1^2 = C_1M^2 - R_{\omega_c}^2 = \text{Pow}(M, \omega_c).$$So, $M$ lies on the radical axis of $\omega_b$ and $\omega_c$, as desired. $\blacksquare$.

Following three steps we can actually easily prove M lies on the radical axis of $\omega_b$ and $\omega_c$.

Great problem! The key observation was that line $FQ$ is tangent to the incircle. After I managed to prove this, the finish was natural! Key Lemma: Line $FQ$ is tangent to the incircle of $ABC$

Now let $U$ and $S$ be intersections of line $FQ$ with line $AB$ and $AC,$ respectively. Further, let $O_b$ and $O_c$ be the centers of $w_b$ and $w_c,$ respectively. From the Key Lemma it follows that $O_b$ is the intersection of line $UI$ with the perpendicular from $B$ to $AB,$ similarly $O_c$ is the intersection of line $VI$ with the perpedicular from $C$ to $AC.$

Claim 3: Quadrilateral $UVBC$ is cyclic.

Claim 4: Points $O_b,O_c$ lies on the circumcircle of $BIC,$ which has center $M.$

To finish let $B'$ and $C'$ be the second intersections of $(BIC)$ with $AB$ and $AC,$ respectively. Combining $O_bB \perp AB, O_cC \perp AC$ with Claim 4, we get that $O_b$ and $O_c$ are the reflections of $B'$ and $C'$ with respect to $M,$ respectively. So we have by Pythagorean Theorem that the power of $M$ with respect to $w_b$ is $$ {MO_b}^2 - {O_bB}^2 = r^2 - (4r^2 - {BB'}^2) = {BB'}^2 -3r^2$$where $r$ is the radius of $(BIC).$ We have an analogous expresion to the power of $M$ with respect to $w_c;$ so that it is sufficient to show that $BB' = CC'.$ Indeed, we have trivially that $AB' = AC$ and $AC' = AB$(say from $\sqrt{bc}$ inversion), so that $BB'=CC'=|AC-AB|$. So done!

Firstly , we'll show that we have $\triangle FIQ \sim \triangle AEI_{a}$ which $I_{a}$ is $A$-excenter of triangle $\triangle ABC$. Suppose that $H$ is the foot of altitude from $A$ to $BC$ and let $AI$ meet line $BC$ at point $D$ , so we have $\triangle AIF \sim \triangle HEA$ and since quadrilateral $QIEI_{a}$ is cyclic , triangles $\triangle QID$ and $\triangle EI_{a}D$ are similar too. So one can see that : $$\frac{AE}{FI}=\frac{HE}{AI} , \frac{QI}{EI_{a}}=\frac{ID}{DE} (I)$$Now since by homothety of incircle and $A$-excircle of triangle $\triangle ABC$ , $AE$ passes trough the antipode of $K$ ( touch point of incircle to $BC$ ) in the incircle , so if we put $H_{a}$ as the midpoint of the altitude $AH$ , then points $E , I , H_{a}$ are collinear and applying Menelaus theorem for these point in triangle $\triangle ADH$ , one can see that : $$\frac{AI}{ID}=\frac{EH}{DE} \implies \frac{AE}{EI_{a}}=\frac{FI}{IQ} , \angle FIQ=\angle AEI_{a}$$So we have $\angle FQI=\angle EI_{a}A=\angle IQD$ and the distance of the point $I$ from lines $BC$ and $FQ$ are equal. As the result , the incircle of triangle $\triangle ABC$ is tangent to $FQ$ and if $FQ$ intersects segments $AB$ and $AC$ at points $X$ , $Y$ respectively , then quadrilateral $BXYC$ is circumscribed and $I$ lies on the bisectors of angles $\angle YXB$ and $\angle XYC$. Now let $O_b$ and $O_c$ be points on the lines $IX$ , $IY$ such that we have $\angle XBO_b=\angle YCO_c=90$ , then obviously these points are centers of circles $\omega_b$ and $\omega_c$. So we have $\angle XQB=2\angle IQD=\angle B-\angle C$ and also $\angle AXY=\angle C$ , so one can see that $\triangle AXY \sim \triangle ABC$ and while $I$ is the $A$-excenter of triangle $\triangle AXY$ we have : $$\angle O_bII_a=\angle AIX=\frac{\angle AYX}{2}=\frac{\angle B}{2}=\angle O_bBI_a$$So quadrilateral $IBI_aO_B$ and similarly $ICI_aO_c$ are cyclic and both points $O_b$ , $O_c$ lies on the circumcircle of triangle $\triangle BIC$ with center $M$ , as the result we have $MO_b=MO_c$ and it's enough to show that equality $BO_b=CO_c$ holds. Let $W_a$ be the touchpoint of $A$-mixtilinear incircle with circumcircle of triangle $\triangle ABC$ and suppose that $QI$ intersects lines $AB$ , $AC$ at points $X'$ , $Y'$ respectively. Now since we know that $\angle CAE =\angle BAW_a$ , then one can see that : $$\frac{EC}{BE}=\frac{BW_{a}}{CW_{a}}.\frac{AC}{AB} (II)$$Now applying Menelaus theorem again for points $Q , X' , Y'$ and $Q , X , Y$ in triangle $\triangle ABC$ , while $AX'=AY'$ we can get : $$\frac{QB}{QC}=\frac{BX'}{CY'} \implies \frac{BX}{CY}=\frac{QB}{QC}.\frac{AX}{AY}=\frac{BX'}{CY'}.\frac{AC}{AB}$$Also a property of mixtilinear shows us that quadrilaterals $BX'IW_a$ and $CY'IW_a$ are cyclic which implies $\triangle BX'W_a \sim \triangle IY'W_a$ and $\triangle CY'W_a \sim \triangle IX'W_a$ , so since $IX'=IY'$ , according to $(II)$ one can see that : $$\frac{BX'}{CY'}=\frac{BW_a}{CW_a} \implies \frac{BX}{CY}=\frac{EC}{BE}$$$$\triangle BXO_b \sim \triangle CEI_a , \triangle CYO_c \sim \triangle BEI_a \implies BO_b=CO_c$$So we're done.

Solved with @levimpcbranco Let $T$ be the contact point of $(ABC)$ and the $A$-Mixtilinear incircle and $N$ the midpoint of arc $BC$ that contains $A$. It's well known ($\sqrt{bc}$ inversion) that $\angle BAT = \angle BCT = \angle CAE \Rightarrow AEB = \angle ACT = \angle ANT = \angle ANI$ since $T$, $I$, $N$ are collinears. Hence $\angle AIF = \angle AEB = \angle ANI \Rightarrow AI^2 = AN \cdot AF$. Let $S$ be the $A$-Chianca Point, i.e, intersection of $(ABC)$ with the circle of diameter $AI$. Since $Q$ is the radical center of $(BIC)$, $(ABC)$ and such circle, $Q$, $S$, $A$ are collinears and by the metric relationships, $AS \cdot AQ = AI^2 = AN \cdot AF \Rightarrow SNFQ$ is cyclic. Hence, $\frac{1}{2} \cdot (B-C) = \angle ASN = \angle AFQ = \angle FQI$. Let $X$, $Y$ be the points where $\omega_b$ and $\omega_c$ touch $FQ$ and $R_b$, $R_c$ the radius of $\omega_b$ and $\omega_c$. Main Claim: $\triangle IBX \sim \triangle ICY$

So $\frac{BX}{CY} = \frac{BI}{CI} = \frac{\sin \frac{C}{2}}{\sin \frac{B}{2}} \Rightarrow 2R_b = \frac{BX}{\sin \frac{C}{2} } = \frac{CY}{\sin \frac{B}{2}} = 2R_c$. Since the radius of the two circles are equal, their radical axis is the perpendicular bisector of $XY$. Let $Z$ be the midpoint of $XY$ and $P$ the midpoint of $BC$. For $MX=MY$, it's suffice that $\angle MZQ = 90^\circ$. In the current spiral similarity, we have $P \rightarrow Z$, so $QZIP$ is cyclic, but also $QIPM$ due to $\angle MPQ = 90^\circ$. Hence, $QZIPM$ is cyclic and $\angle QZM = \angle QPM = 90^\circ$ and we are done.

Config soln: Let the tangent from $Q$ to incircle meet $AB,AC,AM_{BC}$ at $U,V,F'$, it's well known that $BCUV$ is bicentric, so we have $F \equiv F'$ by PoP at $A$, and therefore we also have $FQ$ tangent to incircle, now clearly the $O_b \equiv UI \cap (BIC),O_c \equiv VI \cap (ABC)$, for finishing just let $(BIC)\cap AB,AC = C',B'$ notice $B',M,O_B$ are collinear and by PoP, we're done.