Find the circumcentre of the triangle with ve | vedclass

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(C) Let the circumcentre be $O(x, y)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$OA^2 = OB^2 = OC^2$.Given $A(1, 2)$,$B(

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$(-0.5, 7.0)$

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(C) Let the circumcentre be $O(x, y)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$OA^2 = OB^2 = OC^2$.Given $A(1, 2)$,$B(-2, 2)$,and $C(1, 5)$.$OA^2 = (x - 1)^2 + (y - 2)^2$$OB^2 = (x + 2)^2 + (y - 2)^2$$OC^2 = (x - 1)^2 + (y - 5)^2$Equating $OA^2 = OB^2$:$(x - 1)^2 + (y - 2)^2 = (x + 2)^2 + (y - 2)^2$$(x - 1)^2 = (x + 2)^2$$x^2 - 2x + 1 = x^2 + 4x + 4$$-6x = 3 \implies x = -0.5$Equating $OA^2 = OC^2$:$(x - 1)^2 + (y - 2)^2 = (x - 1)^2 + (y - 5)^2$$(y - 2)^2 = (y - 5)^2$$y^2 - 4y + 4 = y^2 - 10y + 25$$6y = 21 \implies y = 3.5$Thus,the circumcentre is $(-0.5, 3.5)$ or $\left(-\frac{1}{2}, \frac{7}{2}\right)$.

Find the circumcentre of the triangle with vertices $A(1, 2)$,$B(-2, 2)$,and $C(1, 5)$.

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