A triangle has corners at #(9 ,4 )#, #(2 ,5 )#, and #(3 ,6 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Nov 24, 2016

#Area = (125pi)/8#

Explanation:

I always begin this type of problem by shifting all 3 points so that one of them is the origin. Let's shift #(2, 5)# to the origin:

#(2,5) to (0,0)#
#(9,4) to (7, -1)#
#(3,6) to (1, 1)#

I do this, because I am going use the standard equation of a circle:

#(x - h)^2 + (y - k)^2 = r^2#

to write 3 equations using the points. Because the first point is the origin, equation is simplified to become:

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

We can use equation [1] to substitute #h^2 + k^2# for #r^2# in equations [2] and [3]:

#(7 - h)^2 + (-1 - k)^2 = h^2 + k^2" [4]"#
#(1 - h)^2 + (1 - k)^2 = h^2 + k^2" [5]"#

Expand the left sides of equations [4] and [5]:

#49 - 14h + h^2 + 1 + 2k + k^2 = h^2 + k^2" [6]"#
#1 - 2h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [7]"#

cancel the square terms:

#49 - 14h + cancel(h^2) + 1 + 2k + cancel(k^2) = cancel(h^2) + cancel(k^2)" [6]"#
#1 - 2h + cancel(h^2) + 1 - 2k + cancel(k^2) = cancel(h^2) + cancel(k^2)" [7]"#

Collect the constant terms on the right:

#-14h + 2k = -50" [8]"#
#-2h - 2k = -2" [9]"#

Add equation [8] to equation [9]:

#-16h = -52#

#h = 13/4#

A simplified version of equation [9] will help us find the value k:

#13/4 + k = 1#
#k = -9/4#

Use equation [1] to find the value of #r^2#

#r^2 = (13/4)^2 + (9/4)^2#

#r^2 = 250/16 = 125/8#

The area of the circle is:

#Area = pir^2#

#Area = (125pi)/8#