A triangle has corners at #(8 ,1 )#, #(4 ,5 )#, and #(6 ,7 )#. What is the area of the triangle's circumscribed circle?

The three corners are #A (8,1) B (4,5) and C (6,7)#

Distance between two points #(x_1,y_1) and (x_2,y_2)# is

#D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2#

Side #AB= sqrt ((8-4)^2+(1-5)^2)=sqrt(32) ~~ 5.66#unit

Side #BC= sqrt ((4-6)^2+(5-7)^2)=sqrt(8) ~~2.83#unit

Side #CA= sqrt ((6-8)^2+(7-1)^2)=sqrt(40) ~~ 6.32#unit

Area of Triangle is #A_t = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|#

#A_t = |1/2(8(5−7)+4(7−1)+6(1−5))|# or

#A_t = |1/2(-16+24-24)| = 1/2*16 =8.0# sq.unit.

Radius of circumscribed circle is #R=(AB*BC*CA)/(4*A_t)# or

#R=(sqrt(32)*sqrt(8)*sqrt(40))/(4*8) ~~ 3.16#

Area of triangle's circumscribed circle is

#A_c=pi*R^2=pi*3.16^2~~31.42# sq.unit [Ans]

Refer figure abov.

B is mid point of AB

Coordinates of #B = (8+4)/2, (1+5)/2 = ((6,3)#

Similarly coordinates of #F = (4+6)/2, (5+7)/2 = (5,6)#

Slope of AD #m1 = ((5-1) / (4-8)) = -1#

Slope of perpendicular line through B #m_B= -1/m1 = 1#

Eqn of perpendicular line through B is

#y - 3 = 1 *(x - 6)#

#y - x = -3# #color(red)(Eqn (1))#

Slope of CD #m2 = (7-5) / (6-4) = 1#

Slope of perpendicular line through F #m_F = -1/m2 = -1#

Eqn of perpendicular line through F is

#y - 6 = -1 * (x - 5)#

#y + x = 11# #color(red)(Eqn (2)#

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

#y=4, x = 7# #O(7, 4)#

Radius of circumcircle is distance of O from the vertices A, D or C.

I.e # R = OA = OB = OC#

#OA = R = sqrt((8-7)^2 + (1-4)^2) = 3.1623#

Area of circumcircle # A_C = pi R^2 = pi (3.1623)^2 = color (blue)(31.4164)#