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

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Answer 1 · 2017-07-25T16:18:22

The area of the circumscribed circle #=11.34u^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,

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

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

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

We have #3# equations with #3# unknowns

From #(1)# and #(3)#, we get

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

#6a+2b=52-18=34#

#3a+b=17#.............#(4)#

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

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

#6a+4b=43#

#6a+4b=43#..............#(5)#

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

#34-2b=43-4b#, #=>#, #2b=9#, #b=9/2#

#3a=17-b=17-9/2=25/2#, #=>#, #a=25/6#

The center of the circle is #=(25/6,9/5)#

#r^2=(3-25/6)^2+(3-9/2)^2=(-7/6)^2+(-3/2)^2#

#=49/36+9/4#

#=130/36=65/18#

The area of the circle is

#A=pi*r^2=pi*65/18=11.34u^2#