Find the circumcentre and circumradius of a triangle with vertices $(3,0), (-1,-6),$ and $(4,-1)$.

(A) Let the vertices be $A(3,0), B(-1,-6),$ and $C(4,-1)$. Let the circumcentre be $O(x,y)$.
Since $O$ is the circumcentre,$OA^2 = OB^2 = OC^2$.
$OA^2 = (x-3)^2 + (y-0)^2 = x^2 - 6x + 9 + y^2$
$OB^2 = (x+1)^2 + (y+6)^2 = x^2 + 2x + 1 + y^2 + 12y + 36 = x^2 + y^2 + 2x + 12y + 37$
$OC^2 = (x-4)^2 + (y+1)^2 = x^2 - 8x + 16 + y^2 + 2y + 1 = x^2 + y^2 - 8x + 2y + 17$
Equating $OA^2 = OB^2$: $x^2 - 6x + 9 + y^2 = x^2 + y^2 + 2x + 12y + 37 \implies -8x - 12y = 28 \implies 2x + 3y = -7$ (Equation $1$).
Equating $OB^2 = OC^2$: $x^2 + y^2 + 2x + 12y + 37 = x^2 + y^2 - 8x + 2y + 17 \implies 10x + 10y = -20 \implies x + y = -2$ (Equation $2$).
Solving Equations $1$ and $2$: From $2$,$x = -2 - y$. Substituting in $1$: $2(-2-y) + 3y = -7 \implies -4 - 2y + 3y = -7 \implies y = -3$.
Then $x = -2 - (-3) = 1$. So,the circumcentre is $(1, -3)$.
The circumradius $R = OA = \sqrt{(1-3)^2 + (-3-0)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$.