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

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

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

And finally, the circle:

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#

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.