A triangle has corners at #(2 ,4 )#, #(6 ,5 )#, and #(4 ,3 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2017-06-21T11:36:46

The area is #=14.84u^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,

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

#(6-a)^2+(5-b)^2=r^2#..........#(2)#

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

We have #3# equations with #3# unknowns

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

#4-4a+a^2+16-8b+b^2=36-12a+a^2+25-10b+b^2#

#8a+2b=41#.............#(4)#

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

#36-12a+a^2+25-10b+b^2=16-8a+a^2+9-6b+b^2#

#4a+4b=36#

#a+b=9#..............#(5)#

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

#(41-2b)/8=9-b#

#41-2b=72-8b#

#6b=72-41#, #=>#, #b=31/6#

#8a=41-2*31/6=41-31/3=92/3#, #=>#, #a=23/6#

The center of the circle is #=(23/6,31/6)#

#r^2=(2-23/6)^2+(4-31/6)^2=(-11/6)^2+(-7/6)^2#

#=170/36#

#=85/18#

The area of the circle is

#A=pi*r^2=85/18*pi=14.84#