Construct a Circle

The perpendicular bisectors of the line segments $\overline{AB}$ and $\overline{BC}$ will always intersect at the center of the circle through the points $A$ , $B$ , and $C.$ The midpoint of $\overline{AB}$ is $\left( \frac{1+5}{2}, \frac{2+8}{2} \right) = (3,5).$ The slope of the line segment $\overline{AB}$ is $\frac{8-2}{5-1} = \frac{3}{2} ,$ so the slope of the line perpendicular to the line through $A$ and $B$ is $-\frac{2}{3}$ . Hence, an equation of the perpendicular bisector of $\overline{AB}$ is $y - 5 = -\frac{2}{3} (x - 3)$ . Simplifying this equation, we get $y = -\frac{2}{3} x + 7.$

Similarly, we can find that an equation of the perpendicular bisector of $\overline{BC}$ is $y - \frac{15}{2} = 5 \left( x - \frac{15}{2} \right),$ which simplifies to $y = 5x - 30$ .

In order to find the point where these two lines intersect, we set $5x - 30$ equal to $-\frac{2}{3} x + 7$ and solve for $x$ ; when we do so, we get $x = \frac{111}{17}$ . We plug in $x = \frac{111}{17}$ into either equation to get $y = \frac{45}{17},$ and so the center of our circle is $(u,v) = \left( \frac{111}{17}, \frac{45}{17} \right)$ .

Therefore, $u+v = \frac{156}{17}$ and $a+b=156+17 = 173$ .

$\text{Using distance formula, we get,}\\ (u-1)^2+(v - 2)^2= (u - 5)^2+(v - 8)^2 = (u-10)^2+(v - 7)^2\\ \implies~5 - 2u - 4v= 89 - 10u - 16v = 149 - 20u - 14v....Subtracting~~5 - 2u - 4v,\\ We~get~0=84 - 8u - 12v =144 - 18u - 10v\\ \therefore ~~21 - 2u - 3v = 0 .........(1)\\ and~~~~~~~~~72 -9u - 5v = 0 .........(2)\\ 3*(2) - 5*(1),~~give ~~~17u = 111.\\ implies~~u = \dfrac {111}{17}\\$ $Substituting~u~in ~(1),~~21 - 2* \dfrac{111}{17} - 3v = 0,~~v = \dfrac{45}{17}.\\ u+v=\dfrac{156}{17}.~~\therefore~~156+17=~~~\large~~\color{#D61F06}{173}$