Let us name the corners #A(7,6), B(1,3) and C(6,5)#
If we can find the Circumcentre, say, #P(x,y)# of #Delta ABC#, then, we are done, because, in that event, the distance #D# btwn. #P# & either of the corners will give us the Circumradius, say #R#, of #DeltaABC#. With the help of #R#, we can get the Area of the Circumcircle.
Now, as regards finding #P#, we know from Geometry, that, it is equidistant from the corners #A,B,C#. Hence,
#PA^2=PB^2=PC^2 (=R^2)#
#PA^2=PC^2#
#rArr(x-7)^2+(y-6)^2=(x-6)^2+(y-5)^2#
#rArr2x+2y=24 rArr x+y=12...........(1)#
#PC^2=PA^2#
#rArr(x-6)^2+(y-5)^2=(x-1)^2+(y-3)^2#
#rArr10x+4y=51.................(2)#
Solving #(1) & (2)#, we get, #P(x,y)=P(1/2,23/2)#
Hence, #R^2=PB^2=1/4+289/4=145/2#, giving,
The Area of the Circumcircle#=pi*R^2=145pi/2~=227.65sq.unit#.
