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

1 Answer
Nov 16, 2016

#Area = (650pi)/196#

Explanation:

Let's begin by moving the triangle to the left 4 and down 6:

#A = (3, -2), B = (0, 0), and C = (2, 1)#

this does not change the size of the triangle or the circumscribed circle.

The standard form for the equation of a circle is:

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

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

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

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

Please notice that equation [2] simplifies to #h^2 + k^2 = r^2#, therefore, we can temporarily eliminate the variable #r^2# by substitute this into equations [1] and [3]:

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

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

#9 - 6h+ h^2 + 4 + 4k + k^2 = h^2 + k^2" [6]"#
#4 - 4h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [7]"#

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

#9 - 6h + 4 + 4k = 0" [6]"#
#4 - 4h + 1 - 2k = 0" [7]"#

Collect the constant terms on right:

#-6h + 4k = -13" [8]"#
#-4h -2k = -5" [9]"#

Multiply equation [9] by 2 and add to equation [8]:

#-14h + 0k = -23#

h = 23/14

Substitute into equation [9]

#-4(23/14) -2k = -5#

#-2k = 22/14#

#k = -11/14#

Use equation [2] to compute #r^2#

#r^2 = (23/14)^2 + (-11/14)^2#

#r^2 = 650/196#

#Area = pir^2#

#Area = (650pi)/196#