While trying to solve this problem the following conjecture came to my mind. Based on the statement conjectured I could solve the problem mentioned. I am unable to verify the statement that I found interesting in itself. Help needed.
The conjecture
Consider the the concurrent distinct lines $\color{blue}{a}$ and $\color{blue}{b}$, the angle bisectors of the triangle $A,B,C$ at $A$ and $B$ respectively. The inscribed circle $\color{blue}{i}$ is given now.
Move the point $A$ on the line $\color{blue}{a}$ and construct $B'$ and $C'$ such that circle $\color{blue}{i}$ remains the inscribed circle and the lines $\color{blue}{a}$ and $\color{blue}{b}$ remain the angle bisectors as before.
The third vertex $C'$ moves on a circle denoted by $\color{red}{?}$. (The center of $\color{red}{?}$ coincides with the center of $\color{blue}{i}$.)
Let $O$ be the center of the incircle, and say that lines $\overleftrightarrow{OA}$ and $\overleftrightarrow{OB}$ make an angle of size $\varphi$. Let $P$, $P^\prime$, and $Q$ be the points of tangency of $\triangle ABC$ with the incircle, as shown; writing $\theta := \angle AOP = \angle AOQ$, and noting that $P^\prime$ is the reflection of $P$ in $\overleftrightarrow{OB}$, we see that $$\angle AOP^\prime = \angle AOP + 2\angle BOP = \theta + 2(\varphi - \theta) = 2\varphi - \theta$$
Therefore, $\angle QOP^\prime$ (the long way around) is given by $$\angle QOP^\prime = \angle QOA + \angle AOP^\prime = \theta + (2\varphi - \theta) = 2\varphi$$ which is a constant in this scenario. Thus, $\square OP^\prime C Q$ has a constant shape as $A$ moves, and we have easily that $|\overline{OC}| = r\operatorname{sec}\varphi$, where $r$ is the radius of the circle.