Let us name the corners A(7,6), B(1,3) and C(6,5)A(7,6),B(1,3)andC(6,5)
If we can find the Circumcentre, say, P(x,y)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.
