A triangle has corners at #(4 ,7 )#, #(3 ,4 )#, and #(8 ,9 )#. What is the area of the triangle's circumscribed circle?
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The area of the circumscribed circle is #=78.54#
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)#
#(8-a)^2+(9-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=64-16a+a^2+81-18b+b^2#
#10a+10b=120#
#a+b=12#..............#(5)#
From equations #(4)# and #(5)#, we get
#12-b+3b=20#
#2b=8#
#b=8/2=4#
#a=12-b=12-4=8#
The center of the circle is #=(8,4)#
#r^2=(4-a)^2+(7-b)^2=(4-8)^2+(7-4)^2#
#=16+9#
#=25#
The area of the circle is
#A=pi*r^2=25*pi=78.54#
#(4-3)^2+(7-4)^2=10#
#(8-3)^2+(9-4)^2=50#
#(8-4)^2+(9-7)^2=20#
#pi r^2 = {pi (10)(50)(20) }/{ 4(10)(50) - (20-10-50)^2} = 25 pi
#
Here's the shortcut.
The circumcircle is just the circle through the three vertices; the triangle almost doesn't matter. Except, miraculously, the circumradius #r# equals the product of the triangle sides #a,b,c# divided by four times the triangle's area #A#.
#r = {abc}/{4A}#
It's much more useful squared, and we're looking for #pi r^2# anyway.
#pi r^2 = {pi a^2 b^2 c^2}/{16A^2}#
The coordinates give the squared distances easily. Archimedes' Theorem relates the squared distances to the triangle area:
# 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2#
So,
#pi r^2 = {pi a^2 b^2 c^2}/{ 4a^2b^2 - (c^2-a^2-b^2)^2}#
We form the squared distances from pairs of points #(4,7),(3,4), (
8,9) #
#a^2=(4-3)^2+(7-4)^2=10#
#b^2=(8-3)^2+(9-4)^2=50#
#c^2=(8-4)^2+(9-7)^2=20#
#pi r^2 = {pi (10)(50)(20) }/{ 4(10)(50) - (20-10-50)^2} = 25 pi
#
