The circle circumscribed on a triangle is the one that passes through the three vertices.
The general equation for a circle with center #(a,b)# and squared radius #k# is
# (x - a)^2 + (y-b)^2 = k #
Substituting our three points,
# (8 - a)^2 + (3 - b)^2 = k #
# (2 - a)^2 + (4- b)^2 = k#
# (7 -a )^2 + (2-b)^2 = k#
# a^2 + b^2 - 16 a - 6 b + 64 + 9 = k #
# a^2 + b^2 - 4a - 8b + 4 + 16 = k #
# a^2 + b^2 - 14 a - 4 b + 49 + 4 = k#
Subtracting pairs,
#-12 a + 2b + 53 = 0#
# -2 a -2 b + 20 = 0#
Adding,
# -14 a + 73 = 0 #
# a = 73/14#
# 12 a + 12 b - 120 = 0 #
#14b - 67 = 0#
#b = 67/14#
# k = (2 - 73/14 )^2 + (4- 67/14)^2 #
#k = 1/14^2 ( (28-73 )^2 + (56-67)^2 ) = 2146/14^2 = 1073/98 #
Usually they pick nicer numbers for these problems. Our circle's area is #A = \pi r^2 = \pi k#
# A = frac { 1073 pi }{98} #