A triangle has corners at #(8 ,3 )#, #(2 ,4 )#, and #(7 ,2 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-04-18T13:08:10

The circle circumscribed on a triangle is the one that passes through the three vertices.

The general equation for a circle with center #(a,b)# and squared radius #k# is

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

Substituting our three points,

# (8 - a)^2 + (3 - b)^2 = k #

# (2 - a)^2 + (4- b)^2 = k#

# (7 -a )^2 + (2-b)^2 = k#

# a^2 + b^2 - 16 a - 6 b + 64 + 9 = k #

# a^2 + b^2 - 4a - 8b + 4 + 16 = k #

# a^2 + b^2 - 14 a - 4 b + 49 + 4 = k#

Subtracting pairs,

#-12 a + 2b + 53 = 0#

# -2 a -2 b + 20 = 0#

Adding,

# -14 a + 73 = 0 #

# a = 73/14#

# 12 a + 12 b - 120 = 0 #

#14b - 67 = 0#

#b = 67/14#

# k = (2 - 73/14 )^2 + (4- 67/14)^2 #

#k = 1/14^2 ( (28-73 )^2 + (56-67)^2 ) = 2146/14^2 = 1073/98 #

Usually they pick nicer numbers for these problems. Our circle's area is #A = \pi r^2 = \pi k#

# A = frac { 1073 pi }{98} #