A triangle has corners at #(3 ,8 )#, #(7 ,9 )#, and #(4 ,6 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Dec 21, 2016

#Area = (170pi)/36#

Explanation:

The standard Cartesian form of the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2" [1]"#

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

Use equation [1] and the 3 points to write 3 equations:

#(3 - h)^2 + (8 - k)^2 = r^2" [2]"#
#(7 - h)^2 + (9 - k)^2 = r^2" [3]"#
#(4 - h)^2 + (6 - k)^2 = r^2" [4]"#

Expand the squares, using the pattern #(a - b) = a^2 - 2ab + b^2#:

#9 - 6h + h^2 + 64 - 16k + k^2 = r^2" [5]"#
#49 - 14h + h^2 + 81 - 18k + k^2 = r^2" [6]"#
#16 - 8h + h^2 + 36 - 12k + k^2 = r^2" [7]"#

Set the left side of equation [5] equal to the left side of equation [6]:

#9 - 6h + h^2 + 64 - 16k + k^2 = 49 - 14h + h^2 + 81 - 18k + k^2" [8]"#

Set the left side of equation [7] equal to the left side of equation [6]:

#16 - 8h + h^2 + 36 - 12k + k^2 = 49 - 14h + h^2 + 81 - 18k + k^2" [9]"#

There #h^2 and k^2# terms on both sides of equations [8] and [9], therefore, they sum to zero:

#9 - 6h + 64 - 16k = 49 - 14h + 81 - 18k" [10]"#
#16 - 8h + 36 - 12k = 49 - 14h + 81 - 18k" [11]"#

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

#- 6h - 16k = - 14h - 18k + 57" [12]"#
#- 8h - 12k = - 14h - 18k + 78" [13]"#

Collect the h terms into a single term on the left:

#8h - 16k = - 18k + 57" [14]"#
#6h - 12k = - 18k + 78" [15]"#

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

#8h + 2k = 57" [16]"#
#6h + 6k = 78" [17]"#

Multiply equation [17] by #-1/3# and add to equation [16]:

#6h = 31#

#h = 31/6#

Substitute #31/6# for h into equation [17]:

#6(31/6) + 6k = 78" [17]"#

#k = 47/6#

Substitute the values for h and k into equation [3}:

#(7 - 31/6)^2 + (9 - 47/6)^2 = r^2#

#r^2 = (42/6 - 31/6)^2 + (54 - 47/6)^2#

#r^2 = 170/36#

#Area = pir^2#

#Area = (170pi)/36#