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(B) The vertices of the triangle are the intersection points of the lines: $x+y=6$ $(1)$,$2x+y=4$ $(2)$,and $x+2y=5$ $(3)$. Solving $(1)$ and $(2)$: $

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$(\frac{17}{2}, \frac{19}{2})$

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(B) The vertices of the triangle are the intersection points of the lines: $x+y=6$ $(1)$,$2x+y=4$ $(2)$,and $x+2y=5$ $(3)$. Solving $(1)$ and $(2)$: $x = -2, y = 8$. Vertex $A = (-2, 8)$. Solving $(2)$ and $(3)$: $x = 1, y = 2$. Vertex $B = (1, 2)$. Solving $(1)$ and $(3)$: $x = 7, y = -1$. Vertex $C = (7, -1)$. Let the centre of the circle be $(a, b)$. The distance from the centre to each vertex is equal (radius $R$). $(a+2)^2 + (b-8)^2 = (a-1)^2 + (b-2)^2 = (a-7)^2 + (b+1)^2$. From $(a+2)^2 + (b-8)^2 = (a-1)^2 + (b-2)^2$: $a^2 + 4a + 4 + b^2 - 16b + 64 = a^2 - 2a + 1 + b^2 - 4b + 4$. $6a - 12b = -63 \implies 2a - 4b = -21$. From $(a-1)^2 + (b-2)^2 = (a-7)^2 + (b+1)^2$: $a^2 - 2a + 1 + b^2 - 4b + 4 = a^2 - 14a + 49 + b^2 + 2b + 1$. $12a - 6b = 45 \implies 4a - 2b = 15$. Solving the system $2a - 4b = -21$ and $4a - 2b = 15$: Multiply the first by $2$: $4a - 8b = -42$. Subtracting: $(4a - 2b) - (4a - 8b) = 15 - (-42) \implies 6b = 57 \implies b = \frac{57}{6} = \frac{19}{2}$. Substituting $b = \frac{19}{2}$ into $4a - 2b = 15$: $4a - 19 = 15 \implies 4a = 34 \implies a = \frac{34}{4} = \frac{17}{2}$. Thus,the centre $(a, b)$ is $(\frac{17}{2}, \frac{19}{2})$.

If $(a, b)$ is the centre of the circle passing through the vertices of the triangle formed by $x+y=6, 2x+y=4$ and $x+2y=5$,then $(a, b)$ is

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