A triangle has corners at #(9 ,5 )#, #(2 ,3 )#, and #(7 ,6 )#. What is the area of the triangle's circumscribed circle?

2 Answers
Oct 15, 2017

#"Area" = 4505/242pi#

Explanation:

The standard Cartesian form for 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.

We can use the points, #(9,5),(2,3),(7,6)#, and equation [1] to write 3 equations:

#(9 -h)^2 + (5-k)^2 = r^2" [2]"#
#(2 -h)^2 + (3-k)^2 = r^2" [3]"#
#(7 -h)^2 + (6-k)^2 = r^2" [4]"#

Expand the squares:

#81 -18h+ h^2 + 25-10k+k^2 = r^2" [2.1]"#
#4 -4h+h^2 + 9 -6k+k^2 = r^2" [3.1]"#
#49 -14h+h^2 + 36-12k+k^2 = r^2" [4.1]"#

Subtract equation [3.1] from equation [2.1]:

#81 -18h+ h^2 + 25-10k+k^2 = r^2#
#ul(-4 +4h-h^2 - 9 +6k-k^2 = -r^2)#
#77-14h-0h^2+16-4h+0k^2 = 0#

Combine like terms:

#93 - 14h - 4k = 0" [5]"#

Subtract equation [3.1] from equation [4.1]:

#49 -14h+h^2 + 36-12k+k^2 = r^2#
#ul(-4 +4h-h^2 - 9 +6k-k^2 = -r^2)#
#45 + 10h + 0h^2 + 27 + 6k + 0k^2 = 0#

Combine like terms:

#72 - 10h -6k = 0" [6]"#

Multiply both sides of equation [5] by #-3/2# and add to equation [6]:

#72 - 10h -6k - 3/2(93 - 14h - 4k) = 0

Distribute the #-3/2#:

#72 - 10h -6k - 279/2 + 21h + 6k = 0

Combine like terms:

#11h - 135/2#

#h = 135/22#

Use equation [6] to find the value of k:

#72 - 10(135/22) -6k = 0" [6]"#

#234/22 - 6k = 0#

#k = 39/22#

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

#(9 -135/22)^2 + (5-39/22)^2 = r^2#

#r^2 = (63/22)^2+(71/22)^2#

#r^2 = 9010/484#

#r^2 = 4505/242#

Because the area of a circle is #pir^2# we only need to multiply by #pi#:

#"Area" = 4505/242pi#

Oct 15, 2017

Circum center (6.1364, 1.7727)
Area of circumcircle 58.5065

Explanation:

Slope of AB m1 = (3-5)/(2-9)=2/7.
Slope of perpendicular at mid point of AB = -1/m1 = -7/2
Midpoint of AB = (9+2)/2, (3+5)/2 = 11/2, 4
Eqn of perpendicular bisector of AB is
#y - 4 = -(7/2)(x - (11/2))#
#14x+ 4y= 93 color (white)(aaa) Eqn (1)#

Slope of BC m2 = (6-3)/(7-2) = 3/5.
Slope of perpendicular at mid point of AB = -1/m1 = -5/3.
Midpoint of BC = (7+2)/2, (6+3)/2 = 9/2, 9/2
Eqn of perpendicular bisector of BC is
#y - 9/2 = (-5/3)(x - (9/2))#
#10x + 6y = 72 color (white)(aaa) Eqn (2)#

Solving Eqns (1), (2)
#x = (135/22), y=( 39/22)#
Circum center (6.1364, 1.7727)

#r^2 = (9-(135/22))^2+(5-(39/22))^2 #r^2=(63^2+71^2)/22^2#

Area of circumcircle #=pi*r^2 = (cancel22*(63^2+71^2))/(7*cancel(22)*22)#
#=58.5065#

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