A triangle has corners at #(5 ,1 )#, #(2 ,9 )#, and #(4 ,3 )#. What is the area of the triangle's circumscribed circle?

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Answer 1 · 2016-12-07T04:25:33

The area of the circle is #A = 1825/2pi#

When I do this type of problem, I always shift the 3 points so that one of them becomes the origin:

#(5,1) to (0,0)#
#(2,9) to (-3,8)#
#(4,3) to (-1, 2)#

#(x - h)^2 + (y - k)^2 = r^2" [1]"#

where #(x, y)# is any point on the circle, #(h, k)# is the center, and r is the radius.

Use equation [1] and the 3 shifted points to write 3 equations:

#(0 - h)^2 + (0 - k)^2 = r^2" [2]"#
#(-3 - h)^2 + (8 - k)^2 = r^2" [3]"#
#(-1 - h)^2 + (2 - k)^2 = r^2" [4]"#

Equation [2] simplifies to a very useful equation:

#h^2 + k^2 = r^2" [5]"#

Substitute the left side of equation [5] into the right sides of equations [3] and [4]:

#(-3 - h)^2 + (8 - k)^2 = h^2 + k^2" [6]"#
#(-1 - h)^2 + (2 - k)^2 = h^2 + k^2" [7]"#

Expand the squares on the left side of equations [6] and [7]:

#9 + 6h + h^2 + 64 - 16k + k^2 = h^2 + k^2" [8]"#
#1 + 2h + h^2 + 4 - 4k + k^2 = h^2 + k^2" [9]"#

#h^2 + k^2# is on both sides of the equations and this sums to zero:

#9 + 6h + 64 - 16k = 0" [10]"#
#1 + 2h + 4 - 4k = 0" [11]"#

Collect the constant terms into a single term on the right:

#6h - 16k = -73" [12]"#
#2h - 4k = -5" [13]"#

Multiply equation [13] by -4 and add to equation [12]:

#-2h = -53#

#h = 53/2#

Substitute #53/2# for h in equation [13};

#2(53/2) - 4k = -5#

#-4k = -58#

#k = 29/2#

Use equation [5] to compute #r^2#:

#r^2 = (53/2)^2 + (29/2)^2#

#r^2 = 2809/4 + 841/4#

#r^2 = 3650/4#

#r^2 = 1825/2#

The area of the circle is:

#A = pir^2#

#A = 1825/2pi#