A triangle has corners at `(4 ,7 )`, `(3 ,4 )`, and `(8 ,9 )`. What is the area of the triangle's circumscribed circle?

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)`

`(8-a)^2+(9-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=64-16a+a^2+81-18b+b^2`

`10a+10b=120`

`a+b=12`..............`(5)`

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

`12-b+3b=20`

`2b=8`

`b=8/2=4`

`a=12-b=12-4=8`

The center of the circle is `=(8,4)`

`r^2=(4-a)^2+(7-b)^2=(4-8)^2+(7-4)^2`

`=16+9`

`=25`

The area of the circle is

`A=pi*r^2=25*pi=78.54`

Here's the shortcut.

The circumcircle is just the circle through the three vertices; the triangle almost doesn't matter. Except, miraculously, the circumradius `r` equals the product of the triangle sides `a,b,c` divided by four times the triangle's area `A`.

`r = {abc}/{4A}`

It's much more useful squared, and we're looking for `pi r^2` anyway.

`pi r^2 = {pi a^2 b^2 c^2}/{16A^2}`

The coordinates give the squared distances easily. Archimedes' Theorem relates the squared distances to the triangle area:

`16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2`

So,

`pi r^2 = {pi a^2 b^2 c^2}/{ 4a^2b^2 - (c^2-a^2-b^2)^2}`

We form the squared distances from pairs of points `(4,7),(3,4), ( 8,9)`

`a^2=(4-3)^2+(7-4)^2=10`

`b^2=(8-3)^2+(9-4)^2=50`

`c^2=(8-4)^2+(9-7)^2=20`

`pi r^2 = {pi (10)(50)(20) }/{ 4(10)(50) - (20-10-50)^2} = 25 pi`