A triangle has corners at #(6 ,8 )#, #(5 ,4 )#, and #(3 ,2 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Nov 15, 2016

#A = (85pi)/2#

Explanation:

The standard form for the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2#

where #(x,y)# is any given point on the circle, #(h,k)# is the center, and r is the radius.

Use the standard form and the three given points to write 3 equations:

#(6 - h)^2 + (8 - k)^2 = r^2" [1]"#
#(5 - h)^2 + (4 - k)^2 = r^2" [2]"#
#(3 - h)^2 + (2 - k)^2 = r^2" [3]"#

Set the left side of equation [1] equal to the left side of equation [2]:

#(6 - h)^2 + (8 - k)^2 = (5 - h)^2 + (4 - k)^2" [4]"#

Set the left side of equation [1] equal to the left side of equation [3]:

#(6 - h)^2 + (8 - k)^2 = (3 - h)^2 + (2 - k)^2" [5]"#

Use the pattern, #(a - b)^2 = a^2 + 3ab + b^2# to expand the squares:

#36 - 12h + h^2 + 64 - 16k + k^2 = 25 - 10h + h^2 + 16 - 8k + k^2" [6]"#
#36 - 12h + h^2 + 64 - 16k + k^2 = 9 - 6h + h^2 + 4 - 4k + k^2" [7]"#

The #k^2 and h^2# terms cancel:

#36 - 12h + 64 - 16k = 25 - 10h + 16 - 8k" [8]"#
#36 - 12h + 64 - 16k = 9 - 6h + 4 - 4k" [9]"#

Collect the constant terms into a single term on the right:

#-12h - 16k = -10h - 8k - 59 " [10]"#
#-12h - 16k = -6h - 4k - 87" [11]"#

Collect all of the h terms into a single term on the right:

#-16k = 2h - 8k - 59 " [12]"#
#-16k = 6h - 4k - 87" [13]"#

Collect all of the k terms into a single term on the left:

#-8k = 2h - 59 " [14]"#
#-12k = 6h - 87" [15]"#

Divide equation [14] by -8 and equation [15] by -12

#k = -1/4h + 59/8 " [16]"#
#k = -1/2h + 87/12" [17]"#

Set the right side of equation [16] equal to the right side of equation [17]:

#-1/4h + 59/8 = -1/2h + 87/12" [18]"#

Solve for h:

#1/4h = 87/12 - 59/8#

#h = 87/3 - 59/2#

#h = -1/2#

Substitute #-1/2# for h into equation [17]:

#k = 1/4 + 87/12#

#k = 15/2#

Substitute the values of h and k into either equation, [1], [2], or [3]. I will use equation [1]:

#(6 - -1/2)^2 + (8 - 15/2)^2 = r^2#

#(13/2)^2 + (1/2)^2 = r^2#

#r^2 = 170/4 = 85/2#

The area of the circle is #pir^2#:

#A = (85pi)/2#