Question

A triangle has corners at `(3 ,7 )`, `(2 ,5 )`, and `(8 ,4 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Area of the triangle's circumscribed circle: `color(green)(2.65 " sq.units")` (approx.)

Explanation:

Perhaps there is a simpler way, but here is how I would do it:

Part 1: Find the Lengths of the Three Sides
Let `a` be the side `(3,7):(2,5)`
`color(white)("XXX")abs(a)=sqrt((2-3)^2+(5-7)^2)=sqrt(1+4)=sqrt(5)`

Let `b` be the side `(2,5):(8,4)`
`color(white)("XXX")abs(b)=sqrt((8-2)^2+(4-5)^2)=sqrt(36+1)=sqrt(37)`

Let `c` be the side `(8,4):(3,7)`
`color(white)("XXX")abs(c)=sqrt((3-8)^2+(7-4)^2)=sqrt(25+9)=sqrt(34)`

Part 2: Find the Perimeter, Semi-Perimeter, and Area of the Triangle
(I will need the help of a calculator here)
Perimeter: `p=sqrt(5)+sqrt(37)+sqrt(34)~~14.1497824026`

Semi-perimeter: `s=p/2~~7.0748912013`

Area of triangle (using Heron's formula):
`color(white)("XXX")"Area_triangle=sqrt(s(s-a)(s-b)(s-c))~~6.5`

Part 3: Find the Radius and Area of the Circumscribed Circle
(more calculator usage)
Radius of Circumscribed Circle: `r=("Area"_triangle)/s=0.9187420435`
(ask as a separate question if you are unfamiliar with this relation).

Area of Circle: `"Area"_circ = pir^2 =2.6517773377`

Answer 2

`A = (6290pi)/676`

Explanation:

The standard form for the equation of a circle is:

`(x - h)^2 + (y - k)^2 = r^2`

where `(h, k)` is the center and `r` is the radius

Because the triangles vertices are points on the circumscribed circle, we can use the 3 points and the standard form to write 3 equations:

`(3 - h)^2 + (7 - k)^2 = r^2` `" [1]"`
`(2 - h)^2 + (5 - k)^2 = r^2` `" [2]"`
`(8 - h)^2 + (4 - k)^2 = r^2` `" [3]"`

We have 3 equations and we can use them to find the values of, h, k, and `r^2`.

Temporarily eliminate `r^2` by setting the left side of equation [1] equal to the left side of equation [2] and the same for left side of equation [3]:

`(3 - h)^2 + (7 - k)^2 = (2 - h)^2 + (5 - k)^2` `" [1 = 2]"`
`(3 - h)^2 + (7 - k)^2 = (8 - h)^2 + (4 - k)^2` `" [1 = 3]"`

Expand the squares, using the pattern, `(a - b)^2 = a^2 - 2ab + b^2`:

`9 -6h + h^2 + 49 -14k + k^2 = 4 -4h + h^2 + 25 -10k + k^2`
`9 -6h + h^2 + 49 -14k + k^2 = 64 -16h + h^2 + 16 -8k + k^2`

Please notice that for every `h^2 and k^2` on the left there is a corresponding one on the right so they cancel:

`9 -6h + 49 -14k = 4 -4h + 25 -10k`
`9 -6h + 49 -14k = 64 -16h + 16 -8k`

Combine all of the constant terms into a single term on the right:

`- 6h - 14k = - 4h- 10k - 29`
`- 6h- 14k =- 16h- 8k + 22`

Combine all of the h terms into a single term on the right:

`- 14k = 2h- 10k - 29`
`- 14k =- 10h- 8k + 22`

Combine all of the k terms into a single term on the left:

`-4k = 2h - 29` `" [4]"`
`-6k = -10h + 22` `" [5]"`

Multiply equation [4] by -3 and equation [5] by 2 and add them:

`12k - 12k = -6h - 20h + 87 + 44`

`0 = -26h + 131`

`h = 131/26`

Substitute `131/26` for h in equation [4]:

`-4k = 2(131/26) - 29`

`-4k = -492/26`

`k = 123/26`

Substitute the values for h and k into equation [3]

`(8 - 131/26)^2 + (4 - 123/26)^2 = r^2`

`(77/26)^2 + (-19/26)^2 = r^2`

`r^2 = 6290/676`

The area of the circle is `A = pir^2`

`A = (6290pi)/676`