A triangle has corners at #(6 ,8 )#, #(7 ,5 )#, and #(3 ,9 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Nov 1, 2016

#A = pir^2 = (25pi)/2#

Explanation:

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#