A triangle has corners at #(9 ,8 )#, #(2 ,3 )#, and #(1 ,4 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2017-06-06T10:03:01

The area is #=64.6#

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+(8-b)^2=r^2#.......#(1)#

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

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

We have #3# equations with #3# unknowns

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

#81-18a+a^2+64-16b+b^2=4-4a+a^2+9-6b+b^2#

#14a+10b=132#

#7a+5b=66#.............#(4)#

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

#4-4a+a^2+9-6b+b^2=1-2a+a^2+16-8b+b^2#

#2a-2b=-4#

#a-b=-2#..............#(5)#

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

#7(b-2)+5b=66#

#12b=80#

#b=80/12=20/3#

#a=b-2=20/3-2=14/3#

The center of the circle is #=(14/3,20/3)#

#r^2=(1-a)^2+(4-b)^2=(1-14/3)^2+(4-20/3)^2#

#=11^2/3^2+8^2/3^2#

#=20.56#

The area of the circle is

#A=pi*r^2=20.56*pi=64.6#