A triangle has corners at #(4 ,4 )#, #(8 ,2 )#, and #(3 ,1 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-06-05T22:07:32

#{325 pi}/49 #

The radius of the circumcircle equals the product of the sides of the triangle divided by four times the area of the triangle.

This is more useful squared:

# r^2 = {a^2 b^2 c^2}/{16 \ text{area}^2} #

Archimedes' Theorem says

# 16 \ text{area}^2 = 4a^2b^2-(c^2-a^2-b^2)^2#

That all we need to do our problem.

# a^2 = (4-8)^2+(4-2)^2=20#

#b^2 = (8-3)^2+(2-1)^2=26#

#c^2=(4-3)^2+(4-1)^2=10 #

#r^2 = {20(26)(10)}/{ 4(10)(20)-(26-20-10)^2} = 325/49 #

I permuted #a^2, b^2# and #c^2#, necessarily permissible in an area formula.

Anyway the area of the circumcircle is of course #pi r^2 = #

#{325 pi}/49 #