Can You Generalize?

Since $AP$ is an internal bisector of $\angle CAB$ ,

$\dfrac{PB}{PC} = \dfrac {AB}{AC} = 2$

Hence, $CP=80$ and $PB=160$ .

Since $AQ$ is an external bisector of $\angle BAC$ ,

$\dfrac{QB}{QC} = \dfrac {AB}{AC} = 2$

Hence, $QB=2QC$ and therefore $QC+240=2QC$ , which means $QC=240$ .

Also, $\angle QAP=90^{\circ}$ , which means $QP$ is the diameter of the circumcircle of $APQ$ . But $QP=QC+PC = 240 + 80=320$ . Thus, the circumradius is $160$ .

$\text{By Angle Bisector Theorem and Ratio proportion Dividendo-Componendo,}\\ \dfrac {PC}{PB}=\dfrac{AC}{AB}=\dfrac{180}{360}=\frac 1 2.\\ \therefore\ \dfrac {PC}{PB+PC}=\frac 1 {1+2}.\ \ \implies\ \color{#3D99F6}{PC=BC*\frac 1 3=80}.\\ \dfrac {QC}{QB}=\dfrac{AC}{AB}=\dfrac{180}{360}=\frac 1 2\\ \therefore\ \dfrac {QC}{QB-QC}=\frac 1 {1-2}.\ \ \implies\ \color{#3D99F6}{QC=BC*\frac 1 1=240}.\\ \text{But internal and external angle bisectors are at right angles. So PQ is the hypotenuse of }\ \Delta\ APQ.\\ Its\ R=\frac 1 2* hypotenuse=\frac 1 2 *(80+240)=\Large\ \ \ \color{#D61F06}{160}.\\ \ \ \ \\ \\ Generalisation:-\\ \text{Using the above method we get,}\\ \color{#3D99F6}{R=\dfrac{a*b*c}{|b^2-c^2|}.}$