A triangle has corners at #(9 ,4 )#, #(7 ,1 )#, and #(3 ,6 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2017-06-13T07:36:44

The area of the circumscribed circle is #=34.6u^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,

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

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

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

We have #3# equations with #3# unknowns

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

#81-18a+a^2+16-8b+b^2=49-14a+a^2+1-2b+b^2#

#4a+6b=47#

#4a+6b=47#.............#(4)#

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

#49-14a+a^2+1-2b+b^2=9-6a+a^2+36-12b+b^2#

#8a-10b=5#

#8a-10b=5#..............#(5)#

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

#94-12b=5+10b#

#22b=89#

#b=89/22#

#4a=47-6*89/22=47-3*89/11=250/11#, #=>#, #a=250/44=125/22#

The center of the circle is #=(125/22,89/22)#

#r^2=(3-125/22)^2+(6-89/22)^2=(59/22)^2+(43/22)^2#

#=5330/484#

#=2665/242#

The area of the circle is

#A=pi*r^2=2665/242*pi=34.6#