Let the coordinates of the circumcentre O of the triangle DEF be (x, y). Circumcentre of a triangle is equidistant from each of the vertices.
1. Distance OD=distance OE, then:
[tex] (x-1)^2+(y-3)^2=(x-8)^2+(y-3)^2 [/tex].
2. Distance OD=distance OE, then:
[tex] (x-1)^2+(y-3)^2=(x-1)^2+(y+5)^2 [/tex].
Solve the system:
[tex] \left\{\begin{array}{l} (x-1)^2+(y-3)^2=(x-8)^2+(y-3)^2 \\ (x-1)^2+(y-3)^2=(x-1)^2+(y+5)^2 \end{array}\right. [/tex],
[tex] \left\{\begin{array}{l} x^2-2x+1+y^2-6y+9=x^2-16x+64+y^2-6y+9 \\ x^2-2x+1+y^2-6y+9=x^2-2x+1+y^2+10y+25 \end{array}\right. [/tex],
[tex] \left\{\begin{array}{l} 14x=63 \\ -16y=16 \end{array}\right. [/tex].
Then
[tex] x=\dfrac{63}{14} =\dfrac{9}{2} =4.5,\\ \\ y=-1 [/tex].
Answer: the coordinates of the circumcenter for ∆DEF are x=4.5, y=-1.
