Question

A triangle has corners at `(9 ,5 )`, `(2 ,3 )`, and `(7 ,6 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`"Area" = 4505/242pi`

Explanation:

The standard Cartesian form for the equation of a circle is:

`(x -h)^2 + (y-k)^2 = r^2" [1]"`

where `(x,y)` is any point on the circle, `(h,k)` is the center point, and `r` is the radius.

We can use the points, `(9,5),(2,3),(7,6)`, and equation [1] to write 3 equations:

`(9 -h)^2 + (5-k)^2 = r^2" [2]"`
`(2 -h)^2 + (3-k)^2 = r^2" [3]"`
`(7 -h)^2 + (6-k)^2 = r^2" [4]"`

Expand the squares:

`81 -18h+ h^2 + 25-10k+k^2 = r^2" [2.1]"`
`4 -4h+h^2 + 9 -6k+k^2 = r^2" [3.1]"`
`49 -14h+h^2 + 36-12k+k^2 = r^2" [4.1]"`

Subtract equation [3.1] from equation [2.1]:

`81 -18h+ h^2 + 25-10k+k^2 = r^2`
`ul(-4 +4h-h^2 - 9 +6k-k^2 = -r^2)`
`77-14h-0h^2+16-4h+0k^2 = 0`

Combine like terms:

`93 - 14h - 4k = 0" [5]"`

Subtract equation [3.1] from equation [4.1]:

`49 -14h+h^2 + 36-12k+k^2 = r^2`
`ul(-4 +4h-h^2 - 9 +6k-k^2 = -r^2)`
`45 + 10h + 0h^2 + 27 + 6k + 0k^2 = 0`

Combine like terms:

`72 - 10h -6k = 0" [6]"`

Multiply both sides of equation [5] by `-3/2` and add to equation [6]:

#72 - 10h -6k - 3/2(93 - 14h - 4k) = 0

Distribute the `-3/2`:

#72 - 10h -6k - 279/2 + 21h + 6k = 0

Combine like terms:

`11h - 135/2`

`h = 135/22`

Use equation [6] to find the value of k:

`72 - 10(135/22) -6k = 0" [6]"`

`234/22 - 6k = 0`

`k = 39/22`

Use equation [2] to find the value of `r^2`

`(9 -135/22)^2 + (5-39/22)^2 = r^2`

`r^2 = (63/22)^2+(71/22)^2`

`r^2 = 9010/484`

`r^2 = 4505/242`

Because the area of a circle is `pir^2` we only need to multiply by `pi`:

`"Area" = 4505/242pi`

Answer 2

Circum center (6.1364, 1.7727)
Area of circumcircle 58.5065

Explanation:

Slope of AB m1 = (3-5)/(2-9)=2/7.
Slope of perpendicular at mid point of AB = -1/m1 = -7/2
Midpoint of AB = (9+2)/2, (3+5)/2 = 11/2, 4
Eqn of perpendicular bisector of AB is
`y - 4 = -(7/2)(x - (11/2))`
`14x+ 4y= 93 color (white)(aaa) Eqn (1)`

Slope of BC m2 = (6-3)/(7-2) = 3/5.
Slope of perpendicular at mid point of AB = -1/m1 = -5/3.
Midpoint of BC = (7+2)/2, (6+3)/2 = 9/2, 9/2
Eqn of perpendicular bisector of BC is
`y - 9/2 = (-5/3)(x - (9/2))`
`10x + 6y = 72 color (white)(aaa) Eqn (2)`

Solving Eqns (1), (2)
`x = (135/22), y=( 39/22)`
Circum center (6.1364, 1.7727)

#r^2 = (9-(135/22))^2+(5-(39/22))^2 #r^2=(63^2+71^2)/22^2#

Area of circumcircle `=pi*r^2 = (cancel22*(63^2+71^2))/(7*cancel(22)*22)`
`=58.5065`