I was solving some geometry problems before I was stuck onto a problem. The problem says that if tangents from the $A$ point outside the circle are drawn, what would be the distance between the incentre of the triangle formed by $A$ and the points where the tangents touch the cricles. I tried using $PA\cdot PB=PT^2$ but arrived nowhere.
According to your notation, $AP_1 BP_2$ is a cyclic quadrilateral (its vertices lie on a circle having $AB$ as a diameter), hence $\widehat{AP_1 P_2}=\widehat{ABP_2}$ and that is enough to prove that the incenter of $AP_1 P_2$ is just the intersection between $AB$ and the original circle, namely $C$. We also have: $$\frac{1}{2}\widehat{P_1 B C}=\widehat{P_1 P_2 C}=\frac{1}{2}\widehat{P_1 P_2 A}=\frac{1}{2}\left(\frac{\pi}{2}-\widehat{P_2 A B}\right)=\frac{\pi-\widehat{P_2 A P_1}}{4},$$ hence $CP_1 = 2\,BP_1\sin\left(\frac{\widehat{P_1 B C}}{2}\right)$ can be easily described in terms of $AB$ and $r=BP_1$, for instance.