jmc

The circle with center $C$ is the incircle of $△PQR$. So, any segment from a vertex of the triangle to $C$ is an angle bisector.

The sum of the measures of the internal angles of a triangle is $180^{\circ }$, so

$$\begin{array}{rl}\angle QRP& =180^{\circ }-\angle PQR-\angle QPR\\ & =180^{\circ }-63^{\circ }-59^{\circ }\\ & =58^{\circ }.\end{array}$$Since $\overline{RC}$ bisects $\angle QRP$, we have $\angle QRC=\frac{58^{\circ }}{2}=\boxed{29^{\circ }}$.