Let $ABC$ be a triangle in which $\angle BAC = 60^{\circ}$ . Let $P$ (similarly $Q$) be the point of intersection of the bisector of $\angle ABC$(similarly of $\angle ACB$) and the side $AC$(similarly $AB$). Let $r_1$ and $r_2$ be the in-radii of the triangles $ABC$ and $AP Q$, respectively. Determine the circum-radius of $APQ$ in terms of $r_1$ and $r_2$.
1. **Identify the incenter and circumcircle:**
   Let \( I \equiv BP \cap CQ \) be the incenter of \(\triangle ABC\). The circumcircle of \(\triangle APQ\) is denoted by \((U, R_A)\).

2. **Angle properties:**
   Since \(\angle BAC = 60^\circ\), and \(P\) and \(Q\) are points on \(AC\) and \(AB\) respectively, the angle \(\angle PIQ = 120^\circ\). This implies that \(I\) is the midpoint of the arc \(PQ\) of \((U)\).

3. **Midpoint properties:**
   Let \(M \equiv IU \cap PQ\) be the midpoint of \(PQ\). Let \(J\) be the incenter of \(\triangle APQ\).

4. **Projection and cyclic quadrilateral:**
   Let \(E\) be the projection of \(I\) on \(AC\), and \(N\) lies on \(\overline{IE}\) such that \(IM = IN\). Since \(IMEP\) is cyclic due to the right angles at \(M\) and \(E\), we have \(\angle IEM = \angle IPQ = \angle IAQ\).

5. **Similarity of triangles:**
   From the cyclic property, \(\triangle IEM \sim \triangle IAQ\). This gives us the ratio:
   \[
   \frac{IE}{IN} = \frac{IA}{IQ} = \frac{IA}{IJ}
   \]

6. **Parallel lines and radii:**
   Since \(JN \parallel AC\), we have:
   \[
   NE = r_2 = IE - IN = r_1 - \frac{1}{2}R_A
   \]

7. **Solving for \(R_A\):**
   Rearranging the equation, we get:
   \[
   r_2 = r_1 - \frac{1}{2}R_A
   \]
   Solving for \(R_A\), we find:
   \[
   R_A = 2(r_1 - r_2)
   \]

The final answer is \(\boxed{2(r_1 - r_2)}\).