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(B) Let $O(1, 1)$ be the circumcenter. The perpendicular bisector of $AC$ passes through $O(1, 1)$ and the midpoint $P(a, \frac{b+3}{2})$. The slope o

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$9$

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(B) Let $O(1, 1)$ be the circumcenter. The perpendicular bisector of $AC$ passes through $O(1, 1)$ and the midpoint $P(a, \frac{b+3}{2})$. The slope of $AC$ is $\frac{b-3}{a-a}$,which is undefined (vertical line). Thus,the perpendicular bisector is horizontal: $y = \frac{b+3}{2} = 1$ $\Rightarrow b+3=2$ $\Rightarrow b=-1$.Alternatively,from the image,the slope of $OP$ is $\frac{\frac{b+3}{2}-1}{a-1} = -\frac{a-a}{b-3} = 0$. This implies $\frac{b+3}{2} = 1$,so $b=-1$.Using $OQ \perp AB$,where $Q$ is the midpoint of $AB$ $(\frac{a+b}{2}, 4)$,the slope of $AB$ is $\frac{5-3}{b-a} = \frac{2}{b-a}$. The slope of $OQ$ is $\frac{4-1}{\frac{a+b}{2}-1} = \frac{3}{\frac{a+b-2}{2}} = \frac{6}{a+b-2}$.Since $OQ \perp AB$,$(\frac{2}{b-a}) 	imes (\frac{6}{a+b-2}) = -1 \Rightarrow 12 = -(b-a)(a+b-2) = (a-b)(a+b-2)$.Substituting $b=-1$: $12 = (a+1)(a-3) = a^2 - 2a - 3$ $\Rightarrow a^2 - 2a - 15 = 0$ $\Rightarrow (a-5)(a+3) = 0$. So $a=5$ or $a=-3$.Coordinates of $C(a, b)$ are $(5, -1)$ and $(-3, -1)$.The distinct coordinates are $5, -3, -1$. The sum of their absolute values is $|5| + |-3| + |-1| = 5 + 3 + 1 = 9$.

If the circumcenter of the triangle formed by the points $A(a, 3)$,$B(b, 5)$,and $C(a, b)$ is $(1, 1)$,then out of all the possible coordinates of $C$,the sum of the absolute values of the distinct coordinates of $C$ is

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