Question

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

Answer 1

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

Explanation:

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