Let $r_1 = OP = OQ = OT$, $r_2 = PQ = PR = PT$ be radii of circles $\omega_1 \equiv (O, r_1), \omega_2 \equiv (P, r_2) ,$ resp. Let $r_3 = RQ = MR = MP$ be radii of 3 congruent circles $(R, r_3) \cong (P, r_3) \cong (M, r_3).$ From circle $\omega_1 \equiv (O, r_1)$ $\implies$ $\measuredangle TOP = 2 \cdot \measuredangle TQP$. From circle $\omega_2 \equiv (P, r_2)$ $\implies$ $\triangle PTQ$ is P-isosceles $\implies$ its exterior angle at vertex $P$ is equal to $\measuredangle RPQ = 2 \cdot \measuredangle TQP = \measuredangle TOP$ $\implies$ P-, O-isosceles $\triangle RPQ \sim \triangle TOP$ are similar by SAS $\implies$ $\frac{PT}{OT} = \frac{RQ}{RP}$ $\implies$ $r_2^2 = PT \cdot RP = RQ \cdot OT = r_3 \cdot r_1.$ From circles $(P, r_2), (M, r_3)$ $\implies$ $\frac{TN}{PN} = \frac{PT}{PM}$ $\implies$ $TN \cdot r_3 = TN \cdot PM = PT \cdot PN = r_2^2.$ Comparing $\implies$ $TN = r_1 = TO$ $\implies$ circle $(T, TN = r_1)$ passes through $O$.