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

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Answer 1 · 2018-02-03T09:36:11

Area of Circumcircle #A_C = pi 2.7199^2 = color(green)(23.2411)# sq units

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Slope of AB = (y2-y1) / (x2-x1) = (3-6) / (1-5) = 3/4#

Slope of #OM_C# perpendicular to AB is

#m_(M_C) = -1 / (3/4) = -4/3#

Cordinates of #M_c = (5+1)/2, (6 + 3)/2 = (3, 9/2)#

Equation of #OM_C# is

#y - (9/2) = -(4/3) (x - 3)#

#6y - 27 = -8x + 24#

#6y + 8x = 51# Eqn (1)

Slope of BC = (y2-y1) / (x2-x1) = (5-3) / (6-1) = 2/5#

Slope of #OM_A# perpendicular to BC is

#m_(M_A) = -1 / (2/5) = -5/2#

Cordinates of #M_c = (6+1)/2, (5 + 3)/2 = (7/2, 4)#

Equation of #OM_A# is

#y - 4 = -(5/2) (x - (7/2))#

#y - 4 = -(5/4) (2x - 7)#

#4y - 16 = -10x + 35#

#4y + 10x = 51# Eqn (2)

Solving Eqns (1), (2), we get circumcenter coordinates.

#O(51/14, 51/14)#

Radius of circumcenter R is the distance between the circumecenter O and any one of the vertices.

#R = sqrt((5- (51/14))^2 + (6 - (51/14))^2) = 2.7199#

Area of Circumcircle #A_C = pi 2.7199^2 = 23.2411# sq units