To begin with, the vertices of the triangle ABC are as follows;
[tex]\begin{gathered} A=(1,1) \\ B=(1,-3) \\ C=(7,-3) \end{gathered}[/tex]We shall label the center of the triangle (that is the circum-radius) as point O.
That means
[tex]OA=OB=OC[/tex]Note also that the distance between two points on a coordinate grid is given as follows;
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]That means the distance from point OA equals OB and also equals OC.
Therefore;
[tex]\begin{gathered} OA=\sqrt[]{(x-1)^2+(y-1)^2} \\ OA^2=(\sqrt[]{(x-1)^2+(y-1)^2})^2 \\ OA^2=(x-1)^2+(y-1)^2 \\ OA^2=x^2-2x+1+y^2-2y+1 \end{gathered}[/tex]Similarly, distance OB can be calculated as follows;
[tex]\begin{gathered} OB^2=(x-1)^2+(y-\lbrack-3\rbrack)^2 \\ OB^2=(x-1)^2+(y+3)^2 \\ OB^2=x^2-2x+1+y^2+6y+9 \end{gathered}[/tex]Note that
[tex]\begin{gathered} OA=OB \\ \text{Also, } \\ OA^2=OB^2 \\ \text{Therefore;} \\ x^2-2x+1+y^2-2y+1=x^2-2x+1+y^2+6y+9 \\ \text{Collect like terms and we'll now have;} \\ x^2-x^2+y^2-y^2-2x+2x-2y-6y=9+1-1-1 \\ -8y=8 \\ \text{Divide both sides by -8} \\ -\frac{8y}{-8}=\frac{8}{-8} \\ y=-1 \end{gathered}[/tex]We shall now calculate the distance between point OC, as follows;
[tex]\begin{gathered} OC=\sqrt[]{(x-7)^2+(y-\lbrack-3\rbrack)^2} \\ OC^2=(x-7)^2+(y+3)^2 \\ OC^2=x^2-14x+49+y^2+6y+9 \\ \end{gathered}[/tex]Since;
[tex]\begin{gathered} OB=OC \\ \text{And} \\ OB^2=OC^2 \end{gathered}[/tex]We would now have;
[tex]\begin{gathered} x^2-2x+1+y^2+6y+9=x^2-14x+49+y^2+6y+9 \\ \text{Collect all like terms and we'll now have;} \\ x^2-x^2+y^2-y^2-2x+14x+6y-6y=49+9-1-9 \\ 12x=48 \\ \text{Divide both sides by 12;} \\ \frac{12x}{12}=\frac{48}{12} \\ x=4 \end{gathered}[/tex]Therefore, the circumcenter of this triangle with center labelled as point O would be found at the coordinates (4, -1)
ANSWER:
[tex]\text{Circumcenter}=(4,-1)[/tex]