A triangle has corners at #(4 ,7 )#, #(1 ,3 )#, and #(6 ,5 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Apr 21, 2018

#frac{725 pi}{98} #

Explanation:

The circumcised circle is just the circle that has those three points on it. We need the squared radius, so let's solve for the circle's equation.

#(x-a)^2 + (y-b)^2 = r^2#

Each point gives us an equation:

#(4-a)^2+(7-b)^2=r^2#
#(1-a)^2+(3-b)^2=r^2#
#(6-a)^2+(5-b)^2=r^2#

Expanding,

# 16 - 8a + a^2 + 49 - 14b + b^2 = r^2 #
# 1 - 2a + a^2 + 9 - 6b + b^2 = r^2#
# 36 - 12a + a^2 + 25 - 10b + b^2 =r ^2#

Subtracting pairs,

#65 - 10 - 8 a + 2a - 14b + 6b = 0#
# 55= 6a + 8 b#
#65 - 61- 8 a + 12 a - 14 b + 10 b = 0#
# 4 = -4 a + 4 b #
# 1 = -a + b#
# 8 = -8 a + 8 b #

Subtracting,

# 55 - 8 = 6a - -8a #
#a = 47/14#
# b=a+1=61/14#

What we're really after is #r^2#. We substitute any one of the points:

#r^2 = (6 - 47/14)^2 + (5-61/14)^2#
# = frac{1}{14^2} ( ( 6(14)-47)^2 + (5(14) - 61)^2 ) #

#r^2 = 725/98#

The area we seek is

#A = \pi r^2 = frac{725 pi}{98} #