Question

A triangle has corners at `(9 ,7 )`, `(2 ,1 )`, and `(5 ,4 )`. What is the area of the triangle's circumscribed circle?

Answer 1

The area of the circle is aproximately `3.337,84`, with exact form `2125pi"/"2`.

Explanation:

First things first, let us visualise the points we have:

![https://www.desmos.com/calculator]()

These points may look collinear, but they are not.
Now, let's draw the triangle:


And finally, the circle:
![https://www.desmos.com/calculator]()

We can finally start applying formulae and such.

The area of a circle, denoted `S` in this answer, is `pir^2` where `r` is the radius.

In our case, the radius we search is the radius of the circumscribed triangle, `R`.

But how do we find `R`? Well, we could apply the Law of Sines:

`a/sinA=b/sinB=c/sinC = 2R`

So we must find one of the sides and the sine of its opposite angle, say `c=AB` and `/_C`.

The line `AB` is defined by the points `B(2,1)` and `A(9,7)`. Thus, the measure of `AB` is:

`AB=sqrt((2-9)^2+(1-7)^2)=sqrt(49+36)=sqrt85=c`

So we have found `c`. To find `C`, we must use the Law of cosine, which claims that, in any triangle:

`c^color(red)2=a^color(red)2+b^color(red)2-2abcosC`

and its analogues. We are doing this to find the value of `cos C`, from which we can derive `sin C`.

We got to find `a` and `b` too, now:

`BC=sqrt((2-5)^2+(1-4)^2)=sqrt18=3sqrt2=a`
`AC=sqrt((9-5)^2+(7-4)^2)=5=b`

`:. (sqrt85)^color(red)2=(sqrt18)^color(red)2+5^color(red)2-2sqrt18*5cosC`

`85=18+25-10sqrt18cosC => cos C =-42/(10sqrt18)`

`color(blue)(cosC=-7"/"sqrt50)`

We can see that `cos C` is negative, meaning that `C` is an obtuse angle, as visible in the diagram above.

From the Fundamental property of Trigonometry, we have:

`sin^color(red)2C+cos^color(red)2C=1`
`=> sinC=color(red)+sqrt(1-cos^color(red)2C)`

If you're wondering why `sin C` has to be strictly positive; it's because the sine function is positive in both the `"I"^"st"` and `"II"^"nd"` Quadrants.

`sin C = sqrt(1-49/50)=sqrt(1/50)`

Finally, we can find `R` by the Law of Sines:

`c/sinC = 2R`

`sqrt85/sqrt(1/50)=2R => R=1/2sqrt(4250)=sqrt(2125/2)`

`color(blue)(R = sqrt(2125/2)`

So the area of the circle, `S`, is equal to

`color(red)(S = piR^2 = 2125/2 pi ~~3.337,84`

Answer 2

Points `(9,7),(2,1),(5,4)`

`(9-2)^2+(7-1)^2=85`

`(5-2)^2+(4-1)^2=18`

`(9-5)^2+(7-4)^2=25`

`pi r^2 = {pi (85)(18)(25)}/{ 4(85)(18) - (25-85-18)^2} =(2125 pi)/2`

Explanation:

There's just never any need to write a square root. Let's check the other answer.

I don't know why these questions refer to a triangle or its "corners," that is, its vertices. How about:

What's the area of the circle through `(9,7),(2,1)` and `(5,4)` ?

In this answer of mine we find

`pi r^2 = {pi A B C}/{ 4AB - (C-A-B)^2}`

where `A,B,C` are the squared lengths of the sides of the triangle I just tried to dismiss.

The denominator is actually sixteen times the squared area of the triangle. We get the squared lengths from the coordinates with no fuss: `(9,7),(2,1),(5,4)`

`A=(9-2)^2+(7-1)^2=85`

`B = (5-2)^2+(4-1)^2=18`

`C=(9-5)^2+(7-4)^2=25`

`pi r^2 = {pi (85)(18)(25)}/{ 4(85)(18) - (25-85-18)^2} =(2125 pi)/2`

That looks a bit different from the other, featured answer. (EDIT: It had `{2210π}/23` but it's better now.) Who's right? Alpha knows. It's me!

Just eyeballing the figure in the other answer, `r approx 17` looks like about half the radius I got.