Step-by-Step Solution
Key Concept: The circumcentre lies at the intersection of perpendicular bisectors of the sides, equidistant from all vertices.
The perpendicular bisector of $AC$ is $y + 1 = -2(x - 5)$, which simplifies to $y = 2x - 11$. The perpendicular bisector of $BC$ can be found similarly. The circumcentre is at the intersection of these bisectors, found to be $(4, 3)$. The circumradius $R = \sqrt{(4-1)^2 + (3+1)^2} = 5$, and $a + b + R = 4 + 3 + 5 = 12$.
Correct Answer: 2