A triangle has corners at #(7 ,5 )#, #(2 ,3 )#, and #(1 ,4 )#. What is the area of the triangle's circumscribed circle?

1 Answer
Dec 22, 2016

The area of the circumscribed circle is:

#Area = 2146/196pi#

Explanation:

The standard Cartesian form of the equation of a circle is:

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

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

When I do a problem of this type, I shift all 3 points so that one point is then origin, #(0, 0)#, because this simplifies the problem and it does not change the area of this circle:

#(2,3) to (0,0)#

#(1,4) to (-1,1)#

#(7,5) to (5, 2)#

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

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

Equation [2] simplifies into:

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

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

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

Expand the squares:

#1 + 2h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [8]"#
#25 - 10h + h^2 + 4 - 4k + k^2 = h^2 + k^2" [9]"#

The #h^2 and k^2# terms sum to zero:

#1 + 2h + 1 - 2k = 0" [10]"#
#25 - 10h + 4 - 4k = 0" [11]"#

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

#2h - 2k = -2" [10]"#
#-10h -4k = -29" [11]"#

Multiply equation [10] by -2 and add to equation [11]:

#-14h = -25#

#h = 25/14#

Substitute #25/14 for h in equation [10] and the solve for k:

#2(25/14) - 2k = -2#

#k = 1 + 25/14#

#k = 39/14#

Substitute the values for h and k into equation [5]

#r^2 = (25/14)^2 + (39/14)^2#

#r^2 = 2146/196#

The area of a circle is:

#Area = pir^2#

The area of the circumscribed circle is:

#Area = 2146/196pi#