A triangle has corners at #(4 ,7 )#, #(1 ,4 )#, and #(6 ,2 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-05-12T01:07:05

#{841 pi}/98#

I worked out the general form here.

There's a formula we can just apply, but let's just work from the conclusion:

The squared radius of the circumcircle is the product of the squared sides of the triangle divided by #16(text{area})^2#:

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

It's easier to leave everything squared. Since we need to generate the squared sides anyway, we note Archimedes' Theorem:

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

# r^2 = {a^2b^2 c^2}/{ 4 a^2 b^2 - (c^2 -a ^2 - b^2)^2 } #

#(4,7), (1,4), and (6,2)#

We're free to assign #a,b,c# however we like. I prefer #c# to be the longest side.

#a^2 = (4-1)^2 + (7-4)^2=18#

#b^2 = (4-6)^2+(7-2)^2=29#

#c^2 = (6-1)^2 + (2-4)^2 = 29#

Oh, isosceles.

# r^2 = {(18)(29)(29)}/{ 4 (18)(29) - (18^2) } = 841/98 #

The area of the circle is #pi r^2 # or #{841 pi}/98#