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

1 Answer
Jan 5, 2017

Shift all 3 points so that one is the origin. Use the standard Cartesian equation of a circle and the new points to write 3 equations. Solve the 3 equations for #r^2#. Use #r^2# to compute the area.

Explanation:

Shift all three points so that one is the origin:

#(6,3) - (4,2) = (2,1)#
#(5,8) - (4,2) = (1,6)#
#(4,2) - (4,2) = (0,0)#

Use the standard Cartesian equation of a circle:

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

and the three new points to write 3 equations:

#(2 - h)^2 + (1 - k)^2 = r^2" [1]"#
#(1 - h)^2 + (6 - k)^2 = r^2" [2]"#
#(0 - h)^2 + (0 - k)^2 = r^2" [3]"#

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

#4 - 4h + h^2 + 1 - 2k + k^2 = r^2" [4]"#
#1 - 2h + h^2 + 36 - 12k + k^2 = r^2" [5]"#
#h^2 + k^2 = r^2" [6]"#

Subtract equation [6] from equation [4] and [5]:

#4 - 4h + 1 - 2k = 0" [7]"#
#1 - 2h + 36 - 12k = 0" [8]"#

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

#-4h - 2k = -5" [9]"#
#-2h- 12k = -37" [10]"#

Multiply equation [9] by -6 and add to equation [10]:

#22h = -7#

#h = -7/22#

Substitute #-7/22# for h into equation [10] and solve for k:

#-2(-7/22)- 12k = -37#

#7/11- 12k = -37#

#-12k = -37 -7/11#

#k = 69/22#

Use equation [6], to compute the value of #r^2#

#r^2 = (-7/22)^2 + (69/22)^2#

#r^2 = 4810/484 = 2405/242#

The formula for the area of the circle is

#A = pir^2#

Substitute the value for #r^2#:

#A = (2405pi)/242#