A triangle has corners at #(4 ,7 )#, #(8 ,9 )#, and #(3 ,5 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Nov 25, 2016

#Area = (3665pi)/576#

Explanation:

Let's subtract 3 from every x coordinate and 5 from every y coordinate; this will make the 3rd point the origin.

#(1,2), (5,4) and (0,0)#

We do this because we are going to use the standard form of the equation of a circle:

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

and the 3 points to write 3 equations. This makes the equation for the point that is the origin very simple and useful:

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

Substitute the left side of equation [1] into equations [2] and [3]:

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

Expand the squares:

#1 - 2h+h^2 + 4 - 4k+k^2 = h^2 + k^2" [6]"#
#25 - 10h+h^2 + 16 - 8k+k^2 = h^2 + k^2" [7]"#

The square terms cancel:

#1 - 2h+cancel(h^2) + 4 - 4k+cancel(k^2) = cancel(h^2) + cancel(k^2)" [6]"#
#25 - 10h+cancel(h^2) + 16 - 8k+cancel(k^2) = cancel(h^2) + cancel(k^2)" [7]"#

Remove all of the cancelled terms and collect the constants on the right:

#-2h - 4k = -5" [8]"#
#-10h - 8k = -31" [9]"#

Multiply equation [8] by -2 and add to equation [9]:

#-6h = -21#

#h = 7/3#

Substitute into equation [9]:

#-10(7/3) - 8k = -31#

#k = 23/24#

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

#r^2 = (7/3)^ + (23/24)^2#

#r^2 = (7/3)^ + (23/24)^2#

#r^2 = 3665/576#

The area of the circle is:

#Area = pir^2#

#Area = (3665pi)/576#