Question

A triangle has corners at `(7 ,3 )`, `(2 ,5 )`, and `(1 ,4 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Area `A=1073/98 *pi`=34.39723385" "#square units

Explanation:

Let `A(1, 4)` , `B(7, 3)`, `C(2, 5)`

One solution is to determine the orthocenter `H(x_H, y_H)`, then solve for the radius R

`R=sqrt((x_A-x_H)^2+(y_A-y_H)^2)`

the circumcenter is the intersection of the line segment bisectors of the side AB and side AC which have the midpoints at `(4, 7/2)` and `(3/2, 9/2)` respectively.

the equation of the line segment bisector at AB is

`y-7/2=-1/m_(AB)*(x-4)" "`where `" "m_(AB) = -1/6`

`y-7/2=6*(x-4)`

simplified
`12x-2y=41`

the equation of the line segment bisector at AC is

`y-9/2=-1/m_(AC)*(x-3/2)" "`where `m_(AC)=1`

`y-9/2=-1*(x-3/2)`

simplified

`x+y=6`

simultaneous solution using
`x+y=6` and `12x-2y=41` yields

`H(x_H, y_H)=(53/14, 31/14)`

Solve for R now, using any vertex of the triangle ...
let us choose A(1, 4)

`R=sqrt((1-53/14)^2+(4-31/14)^2)`

`R=sqrt(2146)/14`

it follows ,....the area of the circle with center (53/14, 31/14) radius R (see the orange line segment)

`A=pi*R^2`
`A=pi*(sqrt(2146)/14)^2`
`A=1073/98 *pi`=34.39723385" "#square units

God bless....I hope the explanation is useful.