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

#A (9,3), D (4,6), C (2,4)# are the three points given and the corresponding sides are a, d, c.

Slope of line segment #a_m = (6-4) / (4-2) = 1#

Slope of perpendicular bisector passing through F

#F_m = - (1/ a_m) = -1#

Mid point of DC = F has coordinates

#F ( (4+2)/2, (6+4)/2) = F(3,5)

Equation of FO where O is the circumcenter

#y - 5 = -1 (x - 3)#

#color(red)(y + x = 8)# Eqn (1)

Slope of line segment #c_m = (6-3) / (4-9) = -(3/5)#

Slope of perpendicular bisector passing through B

#B_m = - (1/ c_m) = 5/3#

Mid point of AD = B has coordinates

#B ( (9+4)/2, (3+6)/2) = F(13/2,9/2)

Equation of BO where O is the circumcenter

#y - 9/2 = (5/3) (x - 13/2)#

#2y - 9 = (5/3) (2x - 13)#

#color(red)(6y - 10x = -38)# Eqn (2)

Solving Eqns (1), (2), we get the coordinates of circum centerr O

#color(blue)(43/8, 21/8)#

we can get the radius of the circum-circle by finding the distance of O from any one of the three vertices.

#Radius R_c = OA = sqrt((9-(43/8))^2 + (3-(21/8)^2)) = color(green)(3.6443)#

Area of circum-circle #A_c = pi R_c^2 = pi * (3.6443)^2 = color(brown)(41.7232)#