Let AD be internal angle bisector of ∠BAC. Drop perpendicular from B on AD and produce it to meet AC at B'.
Clearly, $\mathrm{△}\mathrm{AMB}\cong \mathrm{△}{\mathrm{AMB}}^{\mathrm{\prime }}$, so  B'  is image of B in line AD.
As slope of AD is – 3, so slope of line BB'  $=\frac{1}{3}=\tan \mathrm{\theta }\left(\text{ say }\right)$    

∴ Equation of BB' in parametric form is $\frac{\mathrm{x}+1}{\frac{3}{\sqrt{10}}}=\frac{\mathrm{y}-3}{\frac{1}{\sqrt{10}}}=\mathrm{r}$
∴ Co-ordinates of B' are $\left(-1+\frac{3}{\sqrt{10}}\times 2\times \frac{5}{\sqrt{10}},3+\frac{1}{\sqrt{10}}\times 2\times \frac{5}{\sqrt{10}}\right)\equiv (2,4)$

                                           $\left(\text{ As distance of line }\mathrm{AD}\text{ from }\mathrm{B}\text{ is }\frac{5}{\sqrt{10}}\right)$

Clearly, equation of  AC  is   y =  2x        ...(1)
Also, equation of AD is  3x + y  =  5 (given)    ...(2)   
∴ On solving (1) and (2), we get A (1, 2)

$\text{ As }\mathrm{AB}=\sqrt{5},\mathrm{AC}=2\sqrt{5}\text{ and }\mathrm{BC}=5$

So, clearly ΔABC is right angled at A. So, by using sine law, we get

$\frac{\mathrm{a}}{\sin \mathrm{A}}=2\mathrm{R}\Rightarrow \frac{5}{\sin {90}^{\circ }}=2\mathrm{R}\Rightarrow \mathrm{R}=\frac{5}{2}$