Lets consider two circles $ w_{1}$ and $ w_{2}$ ortogonal to $ w$ with centers in points $ A$ and $ B$ respectively. $ I\in w_{1} \cap w_{2}$. $ I$ is exist because line $ AB$ doesn't intersect $ w$. Lets consider inversion in point $ I$. We see $ P'_{n + 1}$ is getting from $ P'_{n}$ by composition of two symmetries: relative to lines $ w'_{1}$ and $ w'_{2}$. $ w'_{1}$ and $ w'_{2}$ are passing through center $ O'$ of $ w'$. $ \angle P'_{k}O'P'_{k + 1} = 2 \angle(w'_{1},w'_{2}) = \alpha$ So $ \angle P'_{k}O'P'_{k + 1}$ is constant. $ \angle P'_{k}O'P'_{0} = k\alpha - 2n\pi = 0$ So $ P_{0} = P_{k}$ for any sequence $ P_{0}$, $ P_{1}$, ...,$ P_{k}$ which is built in such way.

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The solution is very nice, thanks! Could the problem G8 from the IMO 1999 also be solved using inversion: Points A, B, C divide the circumcircle $ w$ of the $ \triangle$ABC into three arcs. Let X be a variable point on the arc AB and $ O_{1}$, $ O_{2}$ be the incenters of the $ \triangle$CAX and $ \triangle$CBX. Prove that the circumcircle of the $ \triangle XO_{1}O_{2}$ intersects $ w$ in a fixed point.

This one(IMO)is too simple just use similarities.

We prove, more generally that the statement is true for $w$ any conic and $AB$ a line not intersecting $w$. Problem: Let $w$ be any conic and $AB$ a line not intersecting $w$. Given a point $P_{0}$ on $w$, define the sequence $P_{0},P_{1},\ldots $ as follows: $P_{n+1}$ is the second intersection with $w$ of the line passing through $B$ and the second intersection of the line $AP_{n}$ with $w$. Prove that for a positive integer $k$, if $P_{0}=P_{k}$ for some choice of $P_{0}$, then $P_{0}=P_{k}$ for any choice of $P_{0}$. Proof: Consider a projective transformation which sends $\overleftrightarrow{AB}$ to the line at infinity. So, $w$ gets sent to a conic which does not intersect the line at infinity, therefore an ellipse. Now, we can take a affine transformation which makes $w$ a circle. Thus, it suffices to prove it for $w$ a circle, and $A$ and $B$ as points at infinity. Let $\ell_A, \ell_B$ be two lines which have infinite points as $A,B$. Now, let $Q_i$ be the point on $w$ such that $Q_iP_i \parallel \ell_A$. Then we have $Q_iP_{i+1} \parallel \ell_B$. This implies that $\angle (P_iQ_i, P_{i+1}Q_{i}) = \angle (\ell_A, \ell_B) = \alpha$ and so $P_{i+1}$ is obtained from $P_i$ by a rotation of $\beta = 2 \alpha$ at the centre of $w$. This immediately implies what we desire, since $P_0 = P_k$ if and only if $k\beta = 2n\pi$ for some integer $n$. $\square$