A triangle has corners at #(4 ,7 )#, #(3 ,4 )#, and #(6 ,2 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2017-06-30T07:57:54

The area of the circumscribed circle is #=24.5u^2#

To calculate the area of the circle, we must calculate the radius #r# of the circle

Let the center of the circle be #O=(a,b)#

Then,

#(4-a)^2+(7-b)^2=r^2#.......#(1)#

#(3-a)^2+(4-b)^2=r^2#..........#(2)#

#(6-a)^2+(2-b)^2=r^2#.........#(3)#

We have #3# equations with #3# unknowns

From #(1)# and #(2)#, we get

#16-8a+a^2+49-14b+b^2=9-6a+a^2+16-8b+b^2#

#2a+6b=40#

#a+3b=20#.............#(4)#

From #(2)# and #(3)#, we get

#9-6a+a^2+16-8b+b^2=36-12a+a^2+4-4b+b^2#

#6a-4b=15#..............#(5)#

From equations #(4)# and #(5)#, we get

#6(20-3b)-4b=15#

#120-18b-4b=15#

#22b=105#, #=>#, #b=105/22#

#a=20-3*105/22=125/22#, #=>#, #a=125/22#

The center of the circle is #=(125/22,105/22)#

#r^2=(3-125/22)^2+(4-105/22)^2=(59/22)^2+(17/22)^2#

#=3770/484#

#=1885/242#

The area of the circle is

#A=pi*r^2=pi*1885/242=24.5#