Question

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

Answer 1

`Area = (145pi)/18`

Explanation:

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.

Before we use equation [1] and the 3 given points to write 3 equations, let's move the points so that one of them is the origin. This will not affect the area of the circumscribed circle; it only moves the circle to a different location:

`(5, 1) to (0, 0)`
`(2, 4) to (-3, 3)`
`(7,2) to (2, 1)`

We write the three equations using the new points:

`(0 - h)^2 + (0 - k)^2 = r^2" [2]"`
`(-3 - h)^2 + (3 - k)^2 = r^2" [3]"`
`(2 - h)^2 + (1 - k)^2 = r^2" [4]"`

Equation [2] simplifies:

`h^2 + k^2 = r^2" [5]"`

Substitute left side of equation [5] into right side of equations [3] and [4]:

`(-3 - h)^2 + (3 - k)^2 = h^2 + k^2" [6]"`
`(2 - h)^2 + (1 - k)^2 = h^2 + k^2" [7]"`

Expand the squares:

`9 +6h + h^2 + 9 - 6k + k^2 = h^2 + k^2" [8]"`
`4 - 4h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [9]"`

The `h^2 and k^2` terms cancel:

`9 +6h + 9 - 6k = 0" [10]"`
`4 - 4h + 1 - 2k = 0" [11]"`

collect the constant terms into an single term on the right:

`6h - 6k = -18" [10]"`
`-4h - 2k = -5" [11]"`

Multiply equation [11] by -3 and add to equation [10]:

`18h = -3`

`h = -1/6`

Substitute `-1/6` for h into equation [11]:

`-4(-1/6) - 2k = -5`

`-2k = -5 - 2/3`

`-2k = -17/3`

`k = 17/6`

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

`r^2 = (-1/6)^2 + (17/6)^2`

`r^2 = 290/36 = 145/18`

`Area = pir^2`

`Area = (145pi)/18`

Answer 2

`≈25.31" square units"`

Explanation:

The area (A) of the circle is `color(red)(bar(ul(|color(white)(2/2)color(black)(A=pir^2)color(white)(2/2)|)))`
To calculate the area we require the radius of the circle.

The `color(blue)"circumcentre"` of the circle is at the intersection of the 3 `color(blue)"perpendicular bisectors"` of the sides of the triangle. The perpendicular bisector, bisects the side of a triangle at right angles.

To find the centre we only require 2 equations of perpendicular bisectors, then solve to find coordinates of the centre.

The distance then from the circumcentre to any of the 3 vertices will provide us with the radius and thus area.
`color(magenta)"----------------------------------------------------------------"`
`color(blue)"Equations of perpendicular bisectors"`

`"using the form " color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))`

`" and "color(blue)"gradient formula "m=(y_2-y_1)/(x_2-x_1)`

`"with " m_(perp)=-1/m`

`"side " (5,1)to(2,4)`

`"mid-point " =[1/2(5+2),1/2(1+4)]=(7/2,5/2)`

`m=(4-1)/(2-5)=3/(-3)=-1rArrm_(perp)=1`

`rArry-5/2=x-7/2rArry=x-1to(1)`

`"side " (5,1)to(7,2)`

`"mid-point "=[1/2(5+7),1/2(1+2)]=(6,3/2)`

`m=(2-1)/(7-5)=1/2rArrm_(perp)=-2`

`rArry-3/2=-2(x-6)rArry=-2x+27/2to(2)`
`color(magenta)"-------------------------------------------------------------------"`

`color(blue)"Finding the circumcentre"`

Using equations (1) and (2) from above, both expressed in terms of y.

`rArrx-1=-2x+27/2rArr3x=29/2rArrx=29/6`

substituting in (1): `y=29/6-1=23/6`

`"Hence circumcentre " =(29/6,23/6)`
`color(magenta)"--------------------------------------------------------------"`

`color(blue)"calculating the radius (r) and area"`

`"r is the distance from " (29/6,23/6)` to one of the 3 vertices.

`"Using " (29/6,23/6)to(2,4)"with the"color(blue)" distance formula"`

`r=sqrt((29/6-2)^2+(23/6-4)^2)`

`=sqrt((17/6)^2+(-1/6)^2)`

`rArrr^2=(17/6)^2+(-1/6)^2=289/36+1/36=290/36`

`"Thus area of circumcircle " =290/36pilarr" exact value"`

`rArr"area " ≈25.31" to 2 decimal places"`