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

1 Answer
Douglas K.
Nov 24, 2016

Area = (125pi)/8Area=125π8

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)(2,5) to the origin:

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

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

(x - h)^2 + (y - k)^2 = r^2(x−h)2+(y−k)2=r2

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]"h2+k2=r2 [1]
(7 - h)^2 + (-1 - k)^2 = r^2" [2]"(7−h)2+(−1−k)2=r2 [2]
(1 - h)^2 + (1 - k)^2 = r^2" [3]"(1−h)2+(1−k)2=r2 [3]

We can use equation [1] to substitute h^2 + k^2h2+k2 for r^2r2 in equations [2] and [3]:

(7 - h)^2 + (-1 - k)^2 = h^2 + k^2" [4]"(7−h)2+(−1−k)2=h2+k2 [4]
(1 - h)^2 + (1 - k)^2 = h^2 + k^2" [5]"(1−h)2+(1−k)2=h2+k2 [5]

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

49 - 14h + h^2 + 1 + 2k + k^2 = h^2 + k^2" [6]"49−14h+h2+1+2k+k2=h2+k2 [6]
1 - 2h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [7]"1−2h+h2+1−2k+k2=h2+k2 [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

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