Let the common tangents to $w_1$ and $w_2$ and $w$ at $A,B$ intersect at $S$. Obviously $AS=BS$. By radical axis we have $P,Q,S$ are collinear. We have $O,O_1,A$ collinear and $O,O_2,B$ collinear ($O_i$ is the center of $w_i$). By homothety we have $x=\angle PAO_1=\angle OAB = \angle OBA$ so $\angle AOB = 180-2x$. $\angle ABS = \angle AOB / 2 = 90-x$ so $\angle ASB = 180-2(90-x)=2x$ so $AOBS$ cyclic and easily $APBS$ cyclic since $\angle APB= \angle APQ + \angle BPQ = 90-x+90-x=180-2x$ so we have $O$ is the midarc $APB$ done.

Let $C_1$ and $C_2$ denote the centers of $w_1$ and $w_2$ respectively. Assuming the second circle is greater, we have that $$\angle PAO =\dfrac{180 - \angle AC_1P}{2}=\dfrac{180 - 2\cdot \angle AQP}{2}=90-\angle AQP$$On the other triangle we have $$\angle PBO =\dfrac{180 - \angle BC_2P}{2}=\dfrac{180 -2(180 - \angle BQP)}{2}=\angle BQP - 90 = 90- \angle AQP$$Therefore ABOP is concyclic. So $\angle BPO=\angle BAO=\dfrac{180 - \angle AOB}{2}=\dfrac{180 - \angle BPA}{2}$ From this it follows that $PO$ is the external angle bisector.

Let $OR$ - diameter of $(ABO)$. $R\in PQ$ by Radical Axis. Also $w_1, w_2$ and $(ABR)$ meet in one point ($P$) by Miquel. With $OR$ being perp bisector of $AB$, result follows.

[asy][asy] pair O = (0,2); pair A = (-3,0); pair B = (3,0); pair Q = (1,0); pair T = (0,-4.5); pair P = (1.37647,1.69411); import graph; size(8cm); pen ttffqq = rgb(0.2,1,0); draw(circle(O, 3.60555), linewidth(1) + red); draw(circle((-1,1.33333), 2.40370), linewidth(1) + red); draw(circle((2,0.66666), 1.20185), linewidth(1) + red); draw(T--A, linewidth(1) + blue); draw(T--B, linewidth(1) + blue); draw(P--T, linewidth(1) + ttffqq); draw(A--P, linewidth(1) + red); draw(B--P, linewidth(1) + red); draw(A--B, linewidth(1) + red); draw(circumcircle(A,B,T), linewidth(1) + orange + dashed); dot("$O$", O, dir((5.372, 14.227))); dot("$A$", A, dir((6.058, 13.729))); dot("$B$", B, dir((6.119, 13.729))); dot("$Q$", Q, dir((5.621, 13.729))); dot("$T$", T, dir((5.372, 11.177))); dot("$P$", P, dir((5.209, 11.877))); [/asy][/asy] We rewrite the problem as follows: Consider a circle $\omega$ with center $O$ and chord $AB$. Let $Q$ be a point on chord $AB$, and let $\omega_1$ be the circle through $A$ and $P$ that is tangent to $\omega$. Define $\omega_2$ similarly. Let $P$ be the intersection of $\omega_1$ and $\omega_2$ such that $P \neq Q$. Show that $OP$ is the external angle bisector of $\angle APB$. Claim: Let the tangents of $\omega$ at $A$ and $B$ intersect at $T$. Then $A$, $O$, $P$, $B$ and $T$ are cyclic. Proof: First, notice that from Radical Axis, $PQ$ passes through $T$. Then, directing angles, \[ \angle TBA = \angle BAT = \angle QAT = \angle QPA = TPA \]Thus $APBT$ is cyclic. Then we just note $\angle OAT = 90 = \angle OBT$. $\square$ In particular, this implies that \[ \angle BPT = \angle BAT = \angle TBA = \angle TPA \]Hence $PQ$ actually bisects $\angle APB$. But we also know $\angle OPQ = \angle OPT = \angle OAT = 90$, which implies the conclusion. $\blacksquare$ This is actually the same configuration as AIME I 2019/15.