My solution : Let $ H \equiv BP \cap \odot (MAP) $ . Let $ E \in CA, R \in AB $ be the points s.t. $ \frac{BR}{AR}=\frac{CM}{PM}=\frac{CE}{AE} $ . Since $ A $ is the Miquel point of complete quadrilateral $ \{ \triangle BPC, MH \} $ , so $ \triangle AHB \sim \triangle AMC \Longrightarrow \triangle AHB \cup R \sim \triangle AMC \cup E $ ( $\because \frac{BR}{AR}=\frac{CE}{AE} $ ) , hence combine $ EM \parallel AP $ ( $\because \frac{CM}{PM}=\frac{CE}{AE} $ ) we get $ \triangle RBH \sim \triangle ECM \sim \triangle ACP $ , so $ \angle HRB=\angle PAC=180^{\circ}-\angle CBP \Longrightarrow H $ lie on a fixed circle $ \Omega $ passing through $ R $ and tangent to $ BC $ at $ B $ . $ (\star) $ From Miquel theorem (for $ \triangle BPC $ and points $ M, B, H $) $ \Longrightarrow \odot (BQH) $ is tangent to $ BC $ at $ B \Longrightarrow Q $ lie on $ \Omega $ (from $ (\star) $) . Q.E.D

Do an inversion around $C$. Then $\triangle CAB$ is fixed after inversion, and $M$ moves on a line parallel to $AB$; $P = AB \cap CM$, and $Q$ is the intersection of $BM$ with the circumcircle of $APC$. By Power of a Point (in directed form), \[ BM \cdot BQ = BP \cdot BA \] so $Q$ is the point on the line $BM$ such that $BQ = c \cdot \frac{BP}{BM}$, where $c = BA$. Now, let $R$ be the point on the line $BM$ such that $BR = c \cdot \frac{BM}{BP}$. We claim that $R$ moves along a line. Indeed, let $C=(0,0)$, $P=(p,1)$, $B=(b,1)$, $M=(hp,h)$, then $M-B=(hp-b,h-1)$ so \[ R = B + \left( \frac{hp-b}{b-p}, \frac{h-1}{b-p} \right) \] so that $R-B$ satisfies $-x+by=h$, which is a line. Now, $R$ is the image of $Q$ under an inversion at $B$ of radius $c$, so $Q$ moves on a circle.

Problem. Let $ABC$ be a triangle inscribed in the circle $(O)$. $P$ is a point on arc $\widehat{BC}$ which does not contain $A$. $M$ is a point divide the segment $AP$ in a constant ratio. The circumcircle of the triangle $MPB$ and $MAC$ intersect again at point $Q$. Prove that $Q$ always lies on a fixed circle when $P$ moves. My solution. Choose point $S$ on $BC$ such that $\dfrac{SB}{SC}=\dfrac{MP}{MA}$. Because $M$ is a point divide the segment $AP$ in a constant ratio so $S$ is fixed point. We will prove that the circumcircle of the triangle $SQC$ is fixed circle. Ideed, easily seen $AC,BP$ and $MQ$ are concurrent at $T$. Now $\angle TBQ=\angle PMQ=\angle QCA$. We deduce $BQCT$ is cyclic so $\angle QCB=\angle QTB$. We have $\dfrac{SC}{BT}=\dfrac{SC}{BC}.\dfrac{BC}{BT}=\dfrac{AM}{AP}.\dfrac{AP}{AT}=\dfrac{AM}{AT}=\dfrac{QC}{QT}$. Thus, $\triangle QCS\sim\triangle QTB$. This implies that $\angle QSC=\angle QCA$ or the circumcircle of the triangle $QSC$ is the tangent to $AC$, but $S,C$ are fixed points so the circumcircle of the triangle $SQC$ is fixed circle. We are done.

After perform the inversion with center $ C $ we get the following equivalent problem : Give three non-collinear points $ A, B, C $ and $ P $ be a point varies on line $ AB $ . Let $ \ell $ be a line parallel to $ AB $ and $ M \equiv CP \cap \ell, Q \equiv BM \cap \odot (AMP) $ ($Q \neq M$) . Prove that $ Q $ lie on a fixed circle when $ P $ moves on line $ AB $ ____________________________________________________________ Proof : Let $ T \equiv CA \cap \ell $ and $ X \in BM $ be the point s.t. $ TX \parallel CB $ . Let $ \infty_{UV} $ be the infinity point on the line passing through $ U, V $ . From Pappus theorem (for $ B $-$ M $-$ X $ and $ \infty_{CP} $-$\infty_{BC} $-$\infty_{AB} $) $ \Longrightarrow AX \parallel CP $ , so $ \angle QAB=\angle QMP=\angle BXA \Longrightarrow \triangle BQA \sim \triangle BAX \Longrightarrow {BA}^2=BQ \cdot BX $ , hence $ Q $ lie on the image of the line $ TX $ under the inversion $ \mathbf{I}(\odot(B,BA)) \Longrightarrow Q $ lie on a fixed circle . Q.E.D

Let $\mathcal{I} : X \mapsto X'$ be the inversion with pole $C$ and arbitrary radius $r.$ It is evident that $\mathcal{I}$ sends $\Gamma$ to the line $A'B'.$ Hence, $P'$ is an arbitrary point on $A'B'.$ Meanwhile, since $Q$ is the second intersection of $\odot AMP$ and $\odot (BMC)$, it follows that $Q'$ is the second intersection of $\odot (A'M'P')$ and $B'M'.$ In order to prove that the locus of $Q$ is a circle, it suffices to show that the locus of $Q'$ is a generalized circle (a circle or a line) not passing through $C'.$ To do so, we use complex numbers. We will denote by $x$ the complex number corresponding to the point $X$ after inversion. WLOG let $B'$ be the origin of the complex plane, and let $B'P'$ be the real axis. Because $CM / CP = \lambda$, it follows under inversion that $C'P' / C'M' = \lambda.$ Thus, $C'M' / M'P' = k$ is fixed for some real number $k \in (1, \infty).$ Let us then set $m = \tfrac{c + kp}{1 + k}$, so that $\overline{m} = \tfrac{\overline{c} + kp}{1 + k}.$ From Power of a Point, we obtain \[BP \cdot BA = BM \cdot BQ \implies |p||a| = |m||q| \implies \frac{|p||a|}{|m|^2} = \frac{|q|}{|m|} = \frac{q}{m}.\] Because $|m|^2 = m\overline{m}$, it follows that \[q = \frac{|p||a|}{\overline{m}} = |a|(1 + k) \cdot \frac{p}{\overline{c} + kp}.\] Now, consider the Mobius Transformation $f(z) = |a|(1 + k) \cdot \tfrac{z}{\overline{c} + kz}.$ It is well-known that Mobius Transformations send generalized circles to generalized circles. Then because the locus of $p$ is a generalized circle (the real axis), the locus of $q = f(p)$ is also a generalized circle. It remains to show that the locus of $q$ does not pass through $c.$ Suppose by way of contradiction that $c = f(p)$ for some real number $p.$ Upon multiplication by $\overline{c} + kp$, it follows that \[|a|(1 + k)p = c\overline{c} + kpc = |c|^2 + kpc.\] Hence, $c \in \mathbb{R}$, since every other term in this expression is real. This implies that $a, b, c$ are collinear on the real axis, which is a contradiction. The proof is complete. $\square$

The problem is simple with inversion. And it is nice without inversion. Here I will present both First, non-inversion: Proof: Let $\odot BQC$ meet $AB$ again at $X$. Note that $MX \parallel AP$. By radical axis theorem, $MQ, AP, BC$ concur at a point, call it $Y$. Let $Z = CX \cap AP$. Hence $\angle MXC = \angle YZP = \angle YQP \implies Y, Z, Q, C$ are concyclic. So, $Q$ is the miquel point of $AXBYCZ$. Take point $Q'$ on $AB$ so that $\dfrac{Q'A}{Q'B} =1- \lambda$. Due to Miquel, we get $QZC \sim QAB$, with $X, Q'$ being corresponding points in similarity $\implies \angle ABC = \angle XQC = \angle Q'QB$. Since $Q'$ is fixed, and $\angle ABC$ is fixed, the locus of $Q$ is a circle through $B, Q'$. Second with inversion: Invert about $C$. We get the following problem: Let $CAB$ be a triangle, $\ell$ be a line parallel to $AB$. Take $P$ on $AB$ and let $CP \cap \ell = M$. Let $BM \cap \odot APM = Q$. Show that the locus of $Q$ is a circle. Proof: Take $Q'$ on $AB$ so that $\dfrac{AQ'}{AB} = 1 - \lambda$, and $Q^*$ on $BM$ so that $\dfrac{MQ^*}{MB} = 1 - \lambda$. Since $BMP \sim BAQ$, and $Q^*, Q'$ are corresponding points in similarity, we get $\angle BQQ' = \angle BPQ^*$. But, $PQ^* \parallel CB$ by ratio so $\angle BQQ' = \angle BPQ^* = \angle ABC$. Since $Q'$ and $\angle ABC$ are fixed, $Q$ moves on a circle through $B, Q'$.

Upon inversion with pole $C$, we obtain the following equivalent problem: Let $\triangle ABC$ be a triangle and let $P$ be a variable point on $AB.$ Fix a line $\ell \parallel AB$, and denote $M \equiv AP \cap \ell.$ Let $Q$ be the second intersection of $BP$ with $\odot (AMP).$ Then as $P$ varies, $Q$ moves on a circle. Proof: Let $R \in AB$ satisfy $\tfrac{BR}{BA} = \tfrac{CP}{CM}$ (note that $R$ is a fixed point) and let $F$ be the point such that $\overline{BRA} \cup Q \sim \overline{CPM} \cup F.$ Let $\mathcal{S}$ be the spiral similarity that sends $\overline{BRA} \cup Q \mapsto \overline{CPM} \cup F$ and let $\theta \equiv \measuredangle(AB, MC) = \measuredangle(QA, FM)$ be the angle of rotation. Note that $\measuredangle(BQ, QA) = \measuredangle(MC, AB) = -\theta$ because $A, P, M, Q$ are concyclic. Therefore, $\measuredangle(BQ, FM) = \measuredangle(BQ, QA) + \measuredangle(QA, FM) = 0 \implies F \in BQ.$ It follows from the similarity that $\measuredangle PCF = \measuredangle RBQ = \measuredangle PBF$, implying that $P, F, B, C$ are concyclic. Therefore, $\measuredangle RQB = \measuredangle PFC = \measuredangle PBC$, which is fixed. Thus, $Q$ lies on a fixed circle $\omega$ passing through $B, R.$ Moreover, $CB$ is tangent to $\omega.$

Let $D$ be a point on segment $\overline {BC}$ such that $BD=\lambda \cdot BA$. We claim that the locus of $Q$ is a circle passes through $B$, $D$ and tangent to $BC$, clearly a fixed circle independent of $P$. Note that since $APQM, BQMC, APBC$ are cyclic quadrilaterals, then by Radical Axis Theorem, lines $AP, BC, MQ$ must concur at a point, call this point be $X$. Using oriented angles we get $$\angle (QB,BC)= \angle (QM, MP)= \angle (QA, AP)$$so the four points $A, Q, B, X$ are concyclic. This gives $\angle (QX,XA)= \angle (QB,BA)= \angle (QB,BD) $. If we can prove that $\angle (QA, AX)= \angle (QD, DB) $, then we are done since then $$ \angle (QB, BC)= \angle (QA, AP)= \angle (QD, DB) $$which implies the circumcircle of $BDQ$ is tangent to $BC$ at $B$. However, to prove $\angle (QA, AX)= \angle (QD, DB) $, it is sufficient to prove that $\triangle QXA \sim \triangle QBD $. We claim that $\displaystyle \frac{AX}{XQ}= \frac{BD}{BQ}$. First note that $\triangle XBQ \sim \triangle XMC$, and $\triangle XPC \sim \triangle XBA $, so $\displaystyle \frac{AB}{PC}= \frac{AX}{XC}$ and $\displaystyle \frac{MC}{XC}=\frac{BQ}{XQ} $. Recall that $D$ is chosen such that $\displaystyle \frac{BD}{BA}= \frac{MC}{MP}$, therefore we have $$\frac{BD}{BA}= \frac{MC}{MP}\iff \frac{BD}{MC}= \frac{AB}{PC}= \frac{AX}{XC}\iff BD= AX \cdot \frac{MC}{XC}= AX\cdot \frac{BQ}{XQ}\iff \frac{AX}{XQ}= \frac{BD}{BQ}$$ So we have $\displaystyle \frac{AX}{XQ}= \frac{BD}{BQ}$. Combining with $\angle (QX, XA)= \angle (QB, BD) $, we conclude that $\triangle QXA \sim \triangle QBD $, as required.

Lemma. Given triangle $ABC$ inscribed in $(O)$, $BC$ is fixed, $A$ moves on arc $BC$. Let $P$ be a point on segment $AC$ such that $\dfrac{CP}{AB}=k$ is constant. Then $P$ moves on fixed circle. Proof. Consider a spiral similarity center $Q$ $\mathcal{S}_Q: [AB]\mapsto [PC].$ Then there exist another spiral similarity center $Q$ $\mathcal{S'}_Q: [AP]\mapsto [BC]$. Since $AP$ meets $BC$ at $C$ then $Q\in (ABC)$ and $Q\in (PCC)$. But $\mathcal{S}_Q: B\mapsto C$ we get $\dfrac{QC}{QB}=k$. This means $Q$ is fixed. Therefore $P$ moves on circle through $Q$ and tangent to $BC$ at $C$. Back to problem. Let $G$ be the intersection of $BQ$ and $(O)$, $K$ be the intersection of $GC$ and $QM$. Since $BQMC$ is cyclic then applying Reim theorem, $PG\parallel QM$. Hence $\dfrac{KC}{GC}=\dfrac{MC}{PC}=\lambda.$ We have $\angle AGK=180^\circ-\angle APM=180^\circ-\angle AGC$ then $A,G,M,Q$ are concyclic. Since $(GQK)$ meets $(GBC)$ at $A$, there exist a spiral similarity center $A$ $\mathcal{S}_A: [BG]\mapsto [CK]$. We obtain $\dfrac{BQ}{CK}=\dfrac{AB}{AC}$. From this, $\dfrac{BQ}{CG}=\dfrac{AB}{AC}\cdot \lambda$. Applying lemma above we get $Q$ moves on a circle through $X$ lying on $(O)$ such that $\dfrac{XB}{XC}=\dfrac{AB}{AC}\cdot \lambda$ and tangent to $BC$ at $B.$

navi_09220114 wrote: Let $D$ be a point on segment $\overline {BC}$ such that $BD=\lambda \cdot BA$. We claim that the locus of $Q$ is a circle passes through $B$, $D$ and tangent to $BC$, clearly a fixed circle independent of $P$ İ did not read your whole solution , but as I understood you found what the fixed circle is. How to guess it?Is there any tactic to find it?

Invert around $C$ gives the following problem (ignore the condition $\lambda\in(0,1)$). Let $C, \ell_1, \ell_2$ be fixed point and two parallel lines in the plane. Let $A, B\in\ell_1$ be fixed points. Let $P$ be a variable point on $\ell_1$ and let $M=CP\cap\ell_2$. Let $Q=BM\cap \odot(AMP)$. Prove that the locus of $Q$ is a fixed circle. Note that $\Delta BMP\sim\Delta BQA$. Let $Q_1$ be a point such that $\Delta BPM\cup C\sim\Delta BQA\cup Q_1$. Let $Q_2$ be a point on $BM$ such that $BQ_1 = BQ_2$. Since $\measuredangle Q_2BQ_1 = -\measuredangle PBC$ which is constant and $\frac{AQ}{QQ_1}=\frac{MP}{PC}$ is contant (ratio are directed), we get that mapping $Q\to Q_2$ is spiral similarity. Therefore it suffices to show that $Q_2$ moves along the fixed circle. Note that $$\frac{BQ_1}{BC}=\frac{BA}{BM} \implies BQ_2\cdot BM=BA\cdot BC$$therefore $Q_2$ lies on image of $\ell_2$ w.r.t. inversion around $B$ with power $BA\cdot BC$ hencw we are done.

Start the same way as the post above. Let $C, \ell_1, \ell_2$ be fixed point and two parallel lines in the plane. Let $A, B\in\ell_1$ be fixed points. Let $P$ be a variable point on $\ell_1$ and let $M=CP\cap\ell_2$. Let $Q=BM\cap \odot(AMP)$. Prove that the locus of $Q$ is a fixed circle. Let $D$ be the point on $AB$ such that $BD = BA\cdot\lambda$. We claim that the locus of $Q$ is the circle $\gamma$ through $D$ tangent to line $BC$. Let $MB$ intersect $\gamma$ again at N. It suffices to show that $N=Q$, or equivalently, $BP \cdot BA=BM \cdot BN$. Define $E$ to be the interesection of $CD$ with $\ell_2$, $F$ to be the second intersection of $BE$ with $\gamma$ and $G$ to be the intersection of BC with $\ell_2$. Note that $\frac{BD}{BN}\cdot\frac{FD}{FN}=(D,N;B,F)\stackrel{B}{=}(\infty_{\ell_2},M;G,E)=\frac{ME}{MG} = \frac{PD}{PB}.$ Note that $\measuredangle{NDF}=\measuredangle{MBE},\measuredangle{DFN}=\measuredangle{DBN}=\measuredangle{EMB}\implies \Delta DNF\sim\Delta BEM$. Thus $BN = \frac{BD \cdot FN \cdot PB}{FD \cdot PD} = \frac{BD \cdot ME \cdot PB}{BM \cdot PD} \implies \frac{BN\cdot BM}{PB} = \frac{BD\cdot ME}{PD} = BA$ since $\frac{BD}{BA}=\frac{PD}{ME} = \lambda$, so we are done.

MarkBcc168 wrote: Invert around $C$ gives the following problem (ignore the condition $\lambda\in(0,1)$). Let $C, \ell_1, \ell_2$ be fixed point and two parallel lines in the plane. Let $A, B\in\ell_1$ be fixed points. Let $P$ be a variable point on $\ell_1$ and let $M=CP\cap\ell_2$. Let $Q=BM\cap \odot(AMP)$. Prove that the locus of $Q$ is a fixed circle. . Suppose that $l_2 \cap CB=F.$ Let $D$ be on $AB$ such that $\frac{BD}{BA}=\lambda \implies$ $$\frac{BD}{BQ}=\frac{BD\cdot BM}{BQ \cdot BM}=\frac{BD\cdot BM}{BA\cdot BP}=\frac{BM}{\frac{1}{\lambda}BP}=\frac{BM}{FM}$$And since $\angle{BMF}=\angle{DBQ},$ we have that $\triangle{BQD} \sim \triangle{MBF}\implies \angle{BQD}=\angle{CFM}=\angle{CBA},$ thus $\odot(BQD)$ is tangent to $BC,$ finishing the problem.

By the radical axis theorem on circles $\odot(ABC), \odot(AMP), \odot(BMC)$, lines $AP, BC, MQ$ concur at a point, say $S$. Then by inversion at $S$ we have $(ABQS)$ cyclic. Let $l$ be the tangent line of $\odot(ABS)$ at $S$. Then $\measuredangle ABS=\measuredangle APC$, so $l$ is parallel to $CP$. Hence $$S(CP, M\infty)=(BA, QS),$$where the left-hand side is fixed(as $\lambda$ is fixed). Now invert the diagram at $B$, and denote the images by an apostrophe. Then we have $$B(BA, QS)=(\infty A',Q'S'),$$so there is a constant $c$ such that $\overrightarrow{A'Q'}=c\overrightarrow{A'S'}$. But as $P$ moves along $\odot(ABC)$, $S$ moves along line $BC$, so $S'$ moves on line $BC'$ as well, thus $Q'$ moves on a fixed line not passing $B$. Hence the trajectory of $Q$ is a fixed circle. We're done.

Let $T$ be the point on side $AB$ such that $\frac{AB}{AT} = \lambda$. We take an arbitrary point $P$ on the circumcircle of $ABC$ and prove that the circumcircle of triangle $BTQ$ is tangent to line $BC$; which indeed concludes the problem because when taking a different point, say $P'$, then the circle $BTQ'$ is also tangent to $BC$, hence they are the same circle. So, by radical axis theorem, we have that $AP,QM,BC$ concur at a point $X$. Now notice that the tangency is equivalent to $\angle{QBC} = \angle{QTB}$. Moreover, $\angle{QBC} = \angle{QMP} = \angle{QAX}$ which implies $XAQB$ is cyclic. Therefore, the problem is equivalent to $\triangle{QTB} \sim \triangle{QAX}$, which is equivalent to $\frac{AX}{BT} = \frac{AX}{\lambda \cdot AB} = \frac{QX}{QB}$. But we have $\frac{QX}{QB} = \frac{CX}{MC} = \frac{CX}{\lambda \cdot CP}$ due to the similarity $\triangle{XQB} \sim \triangle{XCM}$. Hence, we need $\frac{AX}{\lambda \cdot AB} = \frac{CX}{\lambda \cdot CP}$, but this is direct from the similarity $\triangle{XAB} \sim \triangle{XCP}$. Therefore, we are done. $\blacksquare$.

As a fact $M$ varies on circle $\gamma$ homothetic to $\Gamma.$ We invert at $C,$ images are marked by primes. Let $X=A'C\cap \gamma ',$ and we claim that inversion at $B'$ with radius $B'A'$ sends $Q'$ to fixed line $A'\infty_{B'C},$ which clearly completes the proof. Indeed, let $Y=B'M'\cap A'\infty_{B'C}$ so by Pappus on $(A'CX),(M'YB')$ we conclude $A'Y\parallel CM'$ and moreover $$\measuredangle B'A'Y=\measuredangle XM'A'=\measuredangle B'A'M'=\measuredangle A'Q'B'\text{ } \blacksquare$$

Let $R$ be the intersection of $AP$ and $BC$. Note that \[\measuredangle BQR=\measuredangle BCM=\measuredangle BAR\]and also that \[\frac{CM}{CR}=\lambda\cdot \frac{CP}{CR}\implies \frac{QB}{QR}=\lambda\cdot \frac{AB}{AR}\implies (B,R;Q,A)=\lambda.\]Now, instead of varying $P$, we vary $R$ instead. Note that after inversion at $B$ we have that \[\frac{A^*R^*}{Q^*R^*}=\lambda\]therefore the locus of $Q^*$ is a line as well; this immediately implies the conclusion. $\blacksquare$

By radical center, let $BC$, $QM$ and $AP$ concur at $E$. Let $F$ be on $AB$ such that $BF:FA=CM:MP$. We claim that $Q$ lies on the fixed circle through $F$ and tangent to $BC$ at $B$. Note that \[\measuredangle BQE=\measuredangle BCP=\measuredangle BAP\]so $BQAE$ is cyclic. Note that \begin{align*} BF &=FA\cdot \frac{MC}{CP}\\ &=MC\cdot \frac{FA}{CP}\\ &=MC\cdot \frac{AE}{EC}\\ &=AE\cdot \frac{BQ}{QE}\\ \end{align*}so $\triangle BFQ\sim^+ \triangle EAQ$ and so $\measuredangle BFQ=\measuredangle EAQ=\measuredangle EBQ$ so $BE$ is tangent to $(BFQ)$ as desired.

Solved with vEnhance's walkthrough on OTIS \ Invert around $C$, and let $\Gamma^{\ast}$ be the image of $\Gamma$. Let $\ell$ be the line for that is the image of $\Gamma^{\ast}$ under a homothety around $C$ (not $C^{\ast}$ of course) with scale factor $1/\lambda.$ Then, $M$ is the intersection of $\ell$ with $CP^{\ast}$. Invert again around $B^{\ast}$ with radius $B^{\ast}A^{\ast}$. Let the image of $Q^{\ast}$ be $Q^{\ast\ast}.$ Claim: $Q^{\ast\ast}A^{\ast}\parallel M^{\ast}P^{\ast}$. By the inversion, we have $$B^{\ast}Q^{\ast}\cdot B^{\ast}Q^{\ast\ast}=(B^{\ast}A^{\ast})^2.$$Furthermore, by Power of a Point on $B^{\ast}$ and circle $(A^{\ast}P^{\ast}M^{\ast})$, $$B^{\ast}M^{\ast}\cdot B^{\ast}Q^{\ast}=B^{\ast}P^{\ast}\cdot B^{\ast}A^{\ast}$$After plugging in the first equation to the second and rearranging, $$\frac{B^{\ast}M^{\ast}}{B^{\ast}Q^{\ast\ast}}=\frac{B^{\ast}P^{\ast}}{B^{\ast}A^{\ast}}.$$Thus, $Q^{\ast\ast}A^{\ast}\parallel M^{\ast}P^{\ast}.$ Let $T$ be the intersection of $\ell$ with $CA^{\ast}$. By Pappus on $Q^{\ast\ast}M^{\ast}B^{\ast}$ and $C^{\ast}A^{\ast}T$, we have $Q^{\ast\ast}T\parallel B^{\ast}C^{\ast}$. Hence, the locus of $Q^{\ast\ast}$ is the line through $T$ parallel to $B^{\ast}C^{\ast}$. Hence, inverting back, $Q^{\ast}$ is on a circle through $B^{\ast}$, and $Q$ is on a circle, as desired.

Absolute doozy of a problem, wow We $\textbf{INVERT THREE TIMES AAAAAAAAAAAAAA HELP}.$ After each I will write down the inverted problem statement. Firstly invert about $C$. This gives the following problem statement: Quote: Let $\lambda \in (0,1)$ be a real number. Given is non-degenerate triangle $ABC$. Point $P$ is a variable point on line $AB$. Point $M$ is the dilation of point $P$ about $C$ with scale factor $\frac{1}{\lambda}$. Line $MB$ intersects the circumcircle of $AMP$ at point $Q$. Show that the locus of $Q$ as $P$ varies is a circle. Now we can actually just discard $\lambda$; instead we will solve a slightly more general version, where $M$ is the intersection of $PC$ and a fixed line parallel to $AB$. Invert again about $A$. Here is the problem statement: Quote: Let $\omega$ be a circle tangent to line $AB$ at $A$, and $C$ a point not on line $AB$. $P$ is a variable point on line $AB$. The circumcircle of $APC$ intersects $\omega$ at $A$ and another point $M$. Line $PM$ intersects the circumcircle of $AMB$ at point $Q$. Show that the locus of $Q$ as $P$ varies is a circle. Observation: taking $P$ arbitrarily close to $B$ implies that the locus of $Q$ should pass through $B$. Invert one last time about $B$. Here is the final problem statement: Quote: Let $\omega$ be a circle tangent to line $AB$ at $A$, and $C$ a point not on line $AB$. $P$ is a variable point on line $AB$. The circumcircle of $APC$ intersects $\omega$ at $A$ and another point $M$. The circumcircle of $PMB$ intersects line $AM$ at $Q$. Show that the locus of $Q$ is a line. A little bit of work needs to be done. Let line $AC$ intersect $\omega$ again at a point $D$, and let line $AC$ intersect line $BQ$ at point $X$. Now note that \[ \measuredangle MCX = \measuredangle MCA = \measuredangle MPA = \measuredangle MPB = \measuredangle MQB = \measuredangle MQX, \]or $MCQX$ is cyclic. Hence then by radical axes it is clear that $PCBX$ is cyclic. Observe that, from points $A$, $B$, $C$, and $X$ one can construct point $P$, and hence points $M$ and $Q$. Hence we can write a reverse problem statement (which is actually solvable this time i swear) Quote: Let $X$ be a point on line $AC$. Given is a line $\ell$ parallel to $AC$, and line $BX$ intersects $\ell$ at $Q$. The circumcircle of $CBX$ intersects line $AB$ at $P$, and the circumcircles of $PAC$ and $PBQ$ intersect at a point $M \neq P$. Show that the locus of $M$ as $X$ varies is a circle tangent to line $AB$ at $A$. Again observe that $QCXM$ is cyclic and $Q$, $A$, $M$ are collinear. Define by $R$ the point on $AC$ such that $RA / RC = XB / XQ$, where lengths are directed. Let $S$ be the second intersection of the circumcircles of $XAB$ and $XCQ$, i.e. the center of the spiral similarity sending $AC$ to $BQ$. Because of the spiral similarity, $S$ must also be the center of the spiral similarity sending $RA$ to $XB$, or the circle $(SRA)$ is tangent to $AB$. Now notice \[ \measuredangle SRA = \measuredangle SXB = \measuredangle SXQ = \measuredangle SMQ = \measuredangle SMA, \]so $SRAM$ is cyclic. It remains to prove that this circle stays fixed, or $\measuredangle RSA$ remains fixed. Indeed, \[ \measuredangle RSA = \measuredangle XSB = \measuredangle XAB, \]which stays constant. We're done. $\textbf{Remark.}$ Maybe I would've found a shorter solution if I didn't have an aversion to even moderate length computation (the only length thing that I used in here was the $RA/RC = XB/XQ$ thing.) $\textbf{Remark 2.}$ Okay apparently this problem is easily solvable after the first inversion using a similar method to the final problem given here. oops For completeness' sake i'll just provide that solution here.

Invert at $C$; then $M$ lies along a fixed line $\ell \parallel \overline{APB}$ and $Q = (APM) \cap \overline{BM}$. Claim. [Eliminating $Q$] Let $R$ inverse of $Q$ with respect to an inversion at $B$ with radius $\overline{AB}$. Then $\overline{AR} \parallel \overline{CPM}$. Proof. Easy angle chasing. $\blacksquare$ Now set $E = \overline{AC} \cap \ell$. I claim $T \in \overline{E\infty_{BC}}$. This is doable by straight coordinates, thus done.

Change the definition of $M$ to the following equivalent one: fix some circle $\omega$ tangent to $\Gamma$ passing through $C$, and let $M=\overline{CP} \cap \omega \neq C$. Now, invert about $C$, which yields the following problem. Restated problem wrote: Let $\overline{AB}$ and $\ell$ be two (distinct) parallel lines and $C$ be a point not on either. Let $P$ be a variable point on $\overline{AB}$ and $M=\ell \cap \overline{CP}$. Let $Q$ be the second intersection of $\overline{BM}$ with $(AMP)$. Prove that as $P$ varies, $Q$ moves along a fixed circle. In fact, I claim that this fixed circle is some circle tangent to $\overline{BC}$ at $B$. To show this, by the law of sines and the angle equality from tangency it suffices to show that $BQ \propto \sin \angle CBQ$. By power of a point, $BQ\cdot BM=BP\cdot BA$, hence $BQ \propto \frac{BP}{BM}$ since $BA$ is fixed. Let $X=\overline{BC} \cap \ell$; by law of sines on $\triangle MXB$ we have $\frac{MX}{\sin \angle CBQ}=\frac{BM}{\sin \angle BXM} \implies \sin \angle CBQ \propto \frac{MX}{BM}$ since $\angle BXM=\angle(\overline{BC},\ell)$ is fixed. It thus suffices to show that $BP \propto MX$, which is obvious by homothety—the ratio is $d(C,\overline{AB})/d(C,\ell)$. $\blacksquare$ Note: If $x$ and $y$ are variable quantities, $x \propto y$ means $x/y$ is constant. Remark: To guess the locus of $Q$, first note that (either before or after inversion) when $P \to B$ we have $Q \to P$ as well, hence the desired circle passes through $B$. To discover the tangency fact, the fact that the circle passes through $B$ means that upon inversion about $B$ (after inversion about $C$) it maps to a line. I didn't do any work with the doubly-inverted problem, but after I drew the diagram it looked a lot like the line that $Q$ was supposed to lie on was parallel to $\overline{BC}$. To confirm these suspicions, one can consider the singly-inverted problem and make $\overline{BC}$ perpendicular to $\overline{AB}$. Then picking two choices of $P$ that are reflections of each other seems to produce two $Q$ that are reflections of each other over $\overline{AB}$, and a bit of lengths confirm this. Furthermore, since lengths work so well in this special case, I tried to apply them in general as well, resulting in the above solution.

Sol:- Let $\Gamma=(O)$, the homothety at $C$ with ratio $\lambda$ maps $(O)$ to $(O')$.Clearly $M \in (O')$. Introduce the circle $(D)$ passing through $A,B$ tangent to $BC$. Observe that $\Delta ADB \stackrel{+}{\sim} \Delta AOC$ which implies that $A$ is the spiral center sending $C \rightarrow B; (O) \rightarrow (D)$ and say $(O') \rightarrow (D')$. Let $MQ \cap BC=R$. Then, $\measuredangle AQR=\measuredangle AQM=\measuredangle APM=\measuredangle APC=\measuredangle ABC=\measuredangle ABR \implies AQBR$ is concyclic.Observe that $\measuredangle BAR=\measuredangle BQR=\measuredangle BQM=\measuredangle BCM=\measuredangle BCP=\measuredangle BAP \implies A-R-P$ are collinear. Let $MQ$ meet $(O')$ again at $E$ and $BQ$ meet $(O)$ again at $F$.Due to reims theorem on cyclic quads $BQCM,BFCP$ we obtain $Q-R-M-E \parallel PF$.Since $ME \parallel PF$ ,by homothety at $C$ ,we obtain $C-E-F$ collinear. By reims theorem on cyclic quad $APFC$ and parallel lines $PF \parallel RE$ we obtain $AREC$ cyclic. Notice $\measuredangle AQB=\measuredangle ARC=\measuredangle AEC$ and $\measuredangle ABQ=\measuredangle ACF=\measuredangle ACE \implies ABQDD' \stackrel{+}{\sim} ACEOO'$. Since $E$ lies on fixed circle $(O')$ , $Q$ lies on fixed circle $(D')$.

$\sqrt{ab}$ invert at $C$. We have the following problem: Let $P$ be a point on $AB$, and $\ell$ be a line parallel to $AB$, and $AP\cap \ell=M$. Now $Q$ is $AM\cap (BMP)$. We want to show the locus of $Q$ is a circle. If $P$ is $A$, then $Q$ is $A$ so the desired locus passes through $A$. This is also seen to be unique. Hence we claim our locus is tangent to $AC$ at $A$. Consider the ratio $\frac{AQ}{\sin \angle CAM}$. If we show this is fixed then the tangent circle to $AC$ through $A$ passing through $Q$ has fixed radius by extended law of sines, thus is the same circle which $Q$ lies on, as desired. Now, $\frac{CM}{\sin \angle CAM}=\frac{AC}{\sin \angle AMP}$ which means \[\frac{1}{\sin \angle CAM}=\frac{AC}{\frac{\sin \angle AMP \times CM\times AP}{AP}}=\frac{AC}{CM\times AP\times\frac{\sin \angle APC}{AM}}=\frac{AC}{\frac{AQ}{AB}\times CM\times \sin\angle APC}=\frac{AC\times AB}{CM\times \sin\angle APC}\times \frac{1}{AQ}\]By repeated sine law and power of a point at $A$ with respect to $(BMP)$. Hence \[\frac{AQ}{\sin\angle CAM}=\frac{AB\times AC}{CM\times \sin\angle APC}=\frac{c}{CM\times \sin\angle APC}\]Thus we want $CM\times \sin \angle APC$ to be fixed. Yet $CM=\lambda CP$, thus we want $CP\sin\angle APC$ to be fixed. Yet this is obvious as \[\frac{CA}{\sin\angle APC}=\frac{CP}{\sin\angle C}\]so the quantity is just $CA\sin\angle C$ which is fixed.