Question

A triangle has corners at `(4 ,7 )`, `(3 ,4 )`, and `(6 ,2 )`. What is the area of the triangle's circumscribed circle?

Answer 1

The area of the circumscribed circle is `=24.5u^2`

Explanation:

To calculate the area of the circle, we must calculate the radius `r` of the circle

Let the center of the circle be `O=(a,b)`

Then,

`(4-a)^2+(7-b)^2=r^2`.......`(1)`

`(3-a)^2+(4-b)^2=r^2`..........`(2)`

`(6-a)^2+(2-b)^2=r^2`.........`(3)`

We have `3` equations with `3` unknowns

From `(1)` and `(2)`, we get

`16-8a+a^2+49-14b+b^2=9-6a+a^2+16-8b+b^2`

`2a+6b=40`

`a+3b=20`.............`(4)`

From `(2)` and `(3)`, we get

`9-6a+a^2+16-8b+b^2=36-12a+a^2+4-4b+b^2`

`6a-4b=15`..............`(5)`

From equations `(4)` and `(5)`, we get

`6(20-3b)-4b=15`

`120-18b-4b=15`

`22b=105`, `=>`, `b=105/22`

`a=20-3*105/22=125/22`, `=>`, `a=125/22`

The center of the circle is `=(125/22,105/22)`

`r^2=(3-125/22)^2+(4-105/22)^2=(59/22)^2+(17/22)^2`

`=3770/484`

`=1885/242`

The area of the circle is

`A=pi*r^2=pi*1885/242=24.5`