EDITED , I thought $C$ is the tangent point of the two small circles ...

Solution : Let w_1 tangent to w_2 at point X w_2 tangent to w at point D QD intersect PC at point W Well known that W is on (ABCD) and NA*NA = NB*NB = NQ*ND = NP*NC O is circumcenter of (PQX) , O is midpoint of QP O' is circumcenter of PQDC O_1 is circumcenter of (PXC) Easy to see that OO_1O' is perpendicular to NP O_1X is tangent to (PQX) and N is on OX Let (PNX) intersect AB at points P' and P K is midpoint of PC KPOO' is cyclic N' is center of w angle P'XN = P'PN = APC = OO'C = OO'O_1 angle O'O_1X = PNX = PP'X angle XO_1O = XPO = OXP = NP'P angle N'NP' = 90 - NP'P = O_1OX So figure O_1OO'X ~ P'NXN' NN'/NX = NN'/NA = OX/OO' = OP/OO' = sqrt(1/2) tan(QCA) = sqrt(1/2) angle QCA = QCP + 45 colculate answere ... . done

$(O)$ is full circle containing the arc $\omega.$ $O_1, O_2$ are centers of $\omega_1, \omega_2,$ tangent to $(O)$ at $C, D.$ $R \equiv AB \cap O_1O_2$ is external similarity center of $\omega_1, \omega_2.$ By Monge theorem, $C, D, R$ are collinear. $X, Y$ are midpoints of the two arcs $AB$ of $(O),$ $X$ on the opposite side of $AB$ than $\omega_1, \omega_2.$ $(X)$ is circle with center $X$ and radius $XA=XB.$ Inversion in $(X)$ takes $(O)$ to $AB$ $\Longrightarrow$ it takes circles $\omega_1, \omega_2$ and their tangency point $I$ to themselves $\Longrightarrow$ $\omega_1, \omega_2 \perp (X)$ and $O_1IO_2$ is tangent of $(X)$ at $I.$ Moreover, $XI$ is the single common internal tangent of $\omega_1, \omega_2,$ cutting $AB, \omega$ at $U, V$ $\Longrightarrow$ $I$ is incenter of $\triangle ABV$ and $UP = UI = UQ.$ $XP \cdot XC = XI^2 = XQ \cdot XD$ $\Longrightarrow$ $PQDC$ is cyclic with circumcircle $(S) \perp (X),$ cutting it at $E, F.$ Radical axes $CD, AB$ of $(O), (S)$ and $(O), (X)$ meet at their radical center $R$ $\Longrightarrow$ radical axis $EF$ of $(S), (X)$ also goes through $R.$ $XE \perp SE, XF \perp SF$ $\Longrightarrow$ $E, F$ are on circle $(K)$ with diameter $XS,$ cutting $(O)$ at $X$ and again at $Z.$ Radical axes $AB, EF$ of $(O), (X)$ and $(K), (X)$ meet at their radical center $R$ $\Longrightarrow$ radical axis $XZ$ of $(O), (K)$ also goes through $R.$ Circle $(L)$ on diameter $XI,$ cutting $(O)$ at $X$ and again at $Z'$ is tangent to $(X)$ at $I.$ Radical axes $AB, O_1IO_2$ of $(O), (X)$ and $(L), (X)$ meet at their radical center $R$ $\Longrightarrow$ radical axis $XZ'$ of $(O), (L)$ also goes through $R.$ If follows that $Z' \equiv Z$ are identical. $YZ, SZ, IZ \perp XZ$ $\Longrightarrow$ $Z, I, S, Y$ are collinear. Since $XY \parallel US$ (both $\perp AB$), $\tan \angle USQ = \frac{UQ}{US} = \frac{UI}{US} = \frac{XI}{XY} = \sin \frac{\angle ACB}{2} .$ $\angle ACQ = \angle ACP + \angle PCQ = \angle ACX + \angle USQ$ $\Longrightarrow$ $\tan \angle ACQ = \frac{\tan \frac{\angle ACB}{2} + \sin \frac{\angle ACB}{2}}{1-\tan \frac{\angle ACB}{2}\sin \frac{\angle ACB}{2}}.$ When $\angle ACB = 90^\circ$, $\tan \angle ACQ = \frac{1+\frac{1}{\sqrt 2}}{1- \frac{1}{\sqrt 2}}= 3+2\sqrt 2.$

dauhuong68 wrote: nice ! why dont you post it sooner? this topic been here for 1 years, and it is your favor. I probably paid no attention to this problem there are so many good problems on the site

Actually，we can take look this problem to prove $\odot CPE$ pass through a fixed point . Solution： Let $\omega_1$ toch $\omega_2$ at $T$，$CP$、$CQ$ respectively intersect $\omega$ at another points $M$、$E$，$AB\cap ME=F$，$N$ is the midpoint of $PQ$ Well known $M$ is the midpoint of $\widehat {AMB}$，$M$、$N$ both are on the radical axes of $\omega_1$ and $\omega$，so $M$、$N$、$T$ are collinear So $MT=MA=MB$，$NT=NP=NQ$，so we can easy to get $2\angle ATP=2\angle BTQ=\angle MAB=45^\circ$ Now，let $TQ$ intersect $\odot(M,MT)$ at another point $G$ . Cuz $MA^2=MP\cdot MC=ME\cdot MF$，so $CPEF$ is cyclic So $QF\cdot QP=QC\cdot QE=QB\cdot QA=QT\cdot QG$，so $TPGF$ is cyclic，so $GF\perp AB$，$2\angle GAF=2\angle BGF=2\angle BTQ=45^\circ$ So $\triangle GAF \sim \triangle BGF$， and obviously $EF$ is the bisector of $\angle AEB$，so： $\tan{\angle ACQ}=\tan{\angle ABE}=\frac{AF}{BF}=\frac{GF^2}{BF^2}=\cot^2{22.5^\circ}=3+2\sqrt{2}$

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Dear MLs This time it is 'right' angle. M.T.

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