Answer: Circumcentre (25/6, 13/6), radius \(\dfrac{13\sqrt2}{6}\)
Let P(x, y) be the circumcentre: PA = PB = PC.
\(PA^2=PB^2\): \((x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2\)
\[-14x+50-2y=-6x+34-10y\ \Rightarrow\ x-y=2\quad\text{(I)}\]
\(PA^2=PC^2\): \((x-7)^2+(y-1)^2=(x-2)^2+y^2\)
\[-14x-2y+50=-4x+4\ \Rightarrow\ 5x+y=23\quad\text{(II)}\]
Adding: \(6x=25\), \(x=\dfrac{25}{6}\), \(y=x-2=\dfrac{13}{6}\).
\[r^2=\left(7-\tfrac{25}{6}\right)^2+\left(1-\tfrac{13}{6}\right)^2=\tfrac{289+49}{36}=\tfrac{169}{18}\]
\[r=\frac{13}{3\sqrt2}=\frac{13\sqrt2}{6}\]