Question

A triangle has corners at `(6 ,4 )`, `(6 ,5 )`, and `(3 ,3 )`. What is the area of the triangle's circumscribed circle?

Answer 1

The area of the circumscribed circle `=11.34u^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,

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

`(6-a)^2+(5-b)^2=r^2`..........`(2)`

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

We have `3` equations with `3` unknowns

From `(1)` and `(3)`, we get

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

`6a+2b=52-18=34`

`3a+b=17`.............`(4)`

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

`36-12a+a^2+25-10b+b^2=9-6a+a^2+9-6b+b^2`

`6a+4b=43`

`6a+4b=43`..............`(5)`

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

`34-2b=43-4b`, `=>`, `2b=9`, `b=9/2`

`3a=17-b=17-9/2=25/2`, `=>`, `a=25/6`

The center of the circle is `=(25/6,9/5)`

`r^2=(3-25/6)^2+(3-9/2)^2=(-7/6)^2+(-3/2)^2`

`=49/36+9/4`

`=130/36=65/18`

The area of the circle is

`A=pi*r^2=pi*65/18=11.34u^2`