Together with the point #(5, 1)# these points form the vertices of a #4xx4# square with centre #(3, 3)#, which is therefore the centre of the circle through the points too.

In general, given three points #A#, #B# and #C# we can construct perpendicular bisectors of line segments #AB# and #BC#, then find their point of intersection, which will be the centre of a circle through them. This is because points on the perpendicular bisector of #AB# are exactly those points which are equidistant from #A# and #B#. So the intersection of the two perpendicular bisectors will be the unique point which is equidistant from #A#, #B# and #C#.

In our example, let #A=(1, 1)#, #B=(1, 5)# and #C=(5, 5)#

Then the midpoint of #AB# is #(1, 3)#. Since #AB# is vertical, its perpendicular bisector is horizontal through #(1, 3)#, i.e. the line #y=3#

The midpoint of #BC# is #(3, 5)#. Since #BC# is horizontal, its perpendicular bisector is vertical through #(3, 5)#, i.e. the line #x=3#.

The lines #y=3# and #x=3# intersect at the point #(3, 3)# as expected.

graph{((x-3)^100+(y-3)^100-2^100)(y-3)(x-3)((x-3)^2+(y-3)^2-8)((x-3)^2+(y-3)^2-0.024)((x-1)^2+(y-5)^2-0.02)((x-5)^2+(y-5)^2-0.02)((x-1)^2+(y-1)^2-0.02)=0 [-6.75, 13.25, -2.88, 7.12]}