A triangle has corners at (9 ,5 ), (2 ,1 ), and (3 ,6 ). What is the area of the triangle's circumscribed circle?

1 Answer
Jan 6, 2017

Shift ALL the points so that one is the origin. Use the standard Cartesian form for the equation of a circle and the new points to write 3 equations. Use the 3 equations to solve for r^2.

Explanation:

Shift all 3 points so that one of them is the origin:

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

This is the standard Cartesian form for the equation of a circle:

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

Use the new points to write 3 equations:

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

Expand the squares:

h^2 + k^2 = r^2" [5]"
49 - 14h + h^2 + 16 - 8k + k^2 = r^2" [6]"
1 - 2h + h^2 + 25 - 10k + k^2 = r^2" [7]"

Subtract equation [6] from equation [5] and equation [7] from equation [5]:

-49 + 14h - 16 + 8k = 0" [8]"
-1 + 2h - 25 + 10k = 0" [9]"

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

14h + 8k = 65" [10]"
2h + 10k = 26" [11]"

Multiply equation [11] by -7 and add to equation [10]:

-62k = -117

k = 117/62

Substitute the value for k into equation [11] and solve for h:

2h + 1170/62 = 26" [11]"

h = 221/62

Use equation [5] to solve for r^2:

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

r^2 = (221/62)^2 + (117/62)^2

r^2 = 62530/3844 = 31265/1922

The area of the circle is:

A = (31265pi)/1922