The standard equation of a circle is:
#(x - h)^2 + (y - k)^2 = r^2#
where #(h,k)# is the centerpoint and r is the radius.
The circumscribed circle will include the three points of the triangle, therefore, we can use the 3 points to write 3 equations:
#(6 - h)^2 + (8 - k)^2 = r^2# [1]
#(7 - h)^2 + (5 - k)^2 = r^2# [2]
#(3 - h)^2 + (9 - k)^2 = r^2# [3]
We can eliminate the variable r by setting the left side of equation [1] equal to the left side of equation [2] and setting the left side of equation [1] equal to the left side of equation [3]:
#(6 - h)^2 + (8 - k)^2 = (7 - h)^2 + (5 - k)^2 # [1 = 2]
#(6 - h)^2 + (8 - k)^2 = (3 - h)^2 + (9 - k)^2# [1 = 3]
Expand the squares:
#36 - 12h+ h^2 + 64 - 16k+ k^2 = 49 - 14h+ h^2 + 25 - 10k+ k^2 #
#36 - 12h+ h^2 + 64 - 16k+ k^2 = 9 - 6h +h^2 + 81 - 18k + k^2#
The square terms cancel:
#36 - 12h+ cancelh^2 + 64 - 16k+ cancelk^2 = 49 - 14h+ cancelh^2 + 25 - 10k+ cancelk^2 #
#36 - 12h+ cancelh^2 + 64 - 16k+ cancelk^2 = 9 - 6h + cancelh^2 + 81 - 18k + cancelk^2#
#36 - 12h + 64 - 16k = 49 - 14h + 25 - 10k#
#36 - 12h + 64 - 16k = 9 - 6h + 81 - 18k#
Collect the constant terms and the terms containing h on the right and collect the terms containing k on the left:
#-6k = -2h - 26# [4]
# 2k = 6h - 10# [5]
Multiply equation [5] by 3 and add to equation [4]:
# 0 = 16h - 56#
#h = 56/16 = 7/2#
Find the value of k by substituting #7/2# for h into equation [5]:
# 2k = 6(7/2) - 10#
#k = 11/2#
The centerpoint is #(7/2, 11/2)#
Find the value of #r^2# by substituting the center point into equation [1]:
#(6 - 7/2)^2 + (8 - 11/2)^2 = r^2#
#(12/2 - 7/2)^2 + (16 - 11/2)^2 = r^2#
#(5/2)^2 + (5/2)^2 = r^2#
#r^2 = 50/4 = 25/2#
Multiply by #pi#, to find the area of the circle:
#A = pir^2 = (25pi)/2#
