A triangle has corners at #(3 ,5 )#, #(4 ,9 )#, and #(7 ,6 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Dec 1, 2016

The area of the circumscribed circle is:

#A = 5.78pi#

Explanation:

When I do this type problem, I always shift all of the corners so that one of them is the origin. This does not change the area of the circumscribed circle and it makes one of the 3 equations (that we must write) become very simple and useful.

Subtract 3 from every x coordinate and subtract 5 from every y coordinate:

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

Use the standard form of the equation of a circle,

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

, and the 3 shifted points to write three equations:

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

Please observe that equation [1] simplifies to:

#h^2 + k^2 = r^2" [4]"#

This allows us to substitute #h^2 + k^2 # for #r^2# in equations [2] and [3]:

#(1 - h)^2 + (4 - k)^2 = h^2 + k^2" [5]"#
#(4 - h)^2 + (1 - k)^2 = h^2 + k^2" [6]"#

Expand the squares, using the pattern #(a - b)^2 = a^2 - 2ab + b^2:

#1 - 2h + h^2 + 16 - 8k + k^2 = h^2 + k^2" [5]"#
#16 - 8h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [6]"#

#h^2 and k^2# are on both side of the equation, therefore, they sum to zero:

#1 - 2h + 16 - 8k = 0" [7]"#
#16 - 8h + 1 - 2k = 0" [8]"#

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

#-2h - 8k = -17" [7]"#
#-8h - 2k = -17" [8]"#

Multiply equation [8] by -4 and add to equation [7]:

#30h = 51#

#h = 1.7#

Substitute 1.7 for h in equation [8]:

#-8(1.7) - 2k = -17#

#-2k = -3.4#

#k = 1.7#

Use equation [4] to find the value of #r^2#:

#r^2 = 1.7^2 + 1.7^2#

#r^2 = 5.78#

The area of a circle is:

#A = pir^2#

The area of the circumscribed circle is:

#A = 5.78pi#