A triangle has corners at #(5 ,7 )#, #(2 ,1 )#, and #(1 ,6 )#. What is the area of the triangle's circumscribed circle?

2 Answers
Feb 2, 2018

Area of circumcircle #A_c = 35.5878#

Explanation:

enter image source here

#M_3 = (5+2)/2, (7+1)/2 = (7/2, 4)#

Slope of A1A2 = (7-1)/(5-2) = 2#

Slope of perpendicular bisector thro #M_3= -1/2#

Equation of perpendicular bisector through #M_3# is

#y - 4 = (-1/2) (x - (7/2))#

#4y - 16 = -2x + 7#

#4y + 2x = 23# Eqn (1)

#M_1 = (2+1)/2, (1+6)/2 = (3/2, 7/2)#

Slope of A2A3 = (6-1) / (1-2) = -6#

Slope of perpendicular bisector through #M_1 = -1/-6 = 1/6#

Equation of perpendicular bisector through #M_1# is

#y - (7/2) = (1/6) (x - (3/2))#

#12y - 42 = 2x - 3#

#12y - 2x = 39# Eqn (2)

Solving equations (1), (2) we get coordinates of circumcenter O.

#O (15/4, 31/8)#

Circumcircle radius #R_o = sqrt(((15/4)-2)^2 + ((31/8) - 1)^2) = 3.3657#

Area of circumcircle #A_c = pi R_o^2 = pi * (3.3657)^2 = 35.5878#

May 8, 2018

Circumcircle area # = {1105 pi}/98#

Explanation:

I worked out the general case here .

I'm going to ignore the details of the general answer, and state the main result as: The squared radius of the circumcircle equals the product of the squared sides of the triangle divided by sixteen times the squared area of the triangle. Given triangle with sides #a,b,c# and area #A#, the circumradius #r# satisfies

# r^2 = {a^2b^2c^2}/{16A^2} #

It's tempting to take the square root, but experience shows much smoother sailing if we don't. We can get the area from the coordinates using the Shoelace Theorem, or get #16A^2# directly from the squared sides using Archimedes' Theorem:

# 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2#

Since we need the squared sides for the numerator anyway, let's do it this way. We'll label the vertices #A(5,7),## B(2,1),## C(1,6) # and calculate

# a^2 = BC^2 = (2-1)^2+(1-6)^2=1^2+5^2=26#

#b^2 = AC^2 = 4^2+1^2=17#

#c^2 = AB^2 = 3^2+6^2=45#

#16 A^2 = 4(26)(17)-(45-26-17)^2 = 1764#

# r^2 = {a^2b^2c^2}/{16A^2} = { (26)(17)(45) }/ 1764 = 1105/98 #

Circumcircle area # = pi r^2 = {1105 pi}/98#