Question

A triangle has corners at `(3 ,7 )`, `(2 ,9 )`, and `(8 ,4 )`. What is the area of the triangle's circumscribed circle?

Answer 1

area `~~166.216` to 3 decimal places

`color(brown)("The approach by")` https://socratic.org/users/ananda-d `color(brown)("is so much simpler!")`

Explanation:

Set point 1 as `P_1->(x_1,y_1)=(2,9)`
Set point 2 as `P_2->(x_2,y_2)=(3,7)`
Set point 3 as `P_3->(x_3,y_3)=(8,4)`

The perpendicular from `(P_1+P_3)/2` will coincide with the perpendicular from `(P_2+P_3)/2` at the centre of the circle
(circumcentre).

`color(blue)("Consider mid point "P_1 to P_3=P_4)`

Set mid point as `P_4`

`x_4=(2+8)/2=5`
`y_4=(9+4)/2=13/2`

`P_4->(x_4,y_4)=(5,13/2)`

`color(blue)("Equation of line perpendicular to "P_1->P_3)`

Set gradient `P_1->P_3=m= (y_3-y_1)/(x_3-x_1)=(4-9)/(8-2)=-5/6`

Perpendicular gradient `=-1/m = +6/5`

So we have `6/5=(y-13/2)/(x-5)`

`6(x-5)=5(y-13/2)`

`6x-30=5y-65/2`

`6x-30+65/2=5y`

`y=6/5x-30/5+65/(5xx2)`

`y=6/5 x+1/2" "......................Equation(1)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Consider mid point "P_2 to P_3=P_5)`

`x_5=(3+8)/2=11/2`

`y_5=(7+4)/2=11/2`

`P_5->(x_5,y_5)=(11/2,11/2)`

Set gradient `P_2->P_3=m=(y_3-y_2)/(x_3-x_2) =(4-7)/( 8-3)=-3/5`

Perpendicular gradient`= -1/m=+5/3`

So we have `5/3=(y-11/2)/(x-11/2)`

`5x-55/2=3y-33/2`

`5x-11=3y`

`y=5/3x-11/3" ".....Equation(2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the centre of the circle")`

Using simultaneous equations for `Eqn(1) and Eqn(2)` we get:

the centre of the circle is at: `P_c->(x_c,y_c)=(625/70, 785/70)`

From this we can derive the radius and hence the area of the circle.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the area of the circle")`

Using Pythagoras on the two points `P_c" and say "P_3`

Let the radius be `r`

`r=sqrt( (x_c-x_3)^2+(y_c-y_3)^2)`

`r=sqrt((625/70-8)^2+(785/70-4)^2)`

`r=sqrt(52.6904...)`

`r~~7.25881......`

area `=pir^2=166.21589...`

area `~~166.216` to 3 decimal places

Answer 2

`~~166.2` (see below)

Explanation:

If you just need the area of the circumscribed circle, the amount of calculations can be reduced a bit by appealing to a simple consequence of the sine law of triangles.

We usually express this law in the form

`a/sin A = b/sin B = c/sin C`

what is often not mentioned is that this common value is, actually the diameter `d` of the circumscribed circle. For a proof, see the figure below

Here ABC is a triangle whose circumscribed circle is drawn in red. AD is a diameter of the circle. Then `/_C = /_BDA` (angles in the same segment of a circle) and `Delta ABD` is right-angled at B (angle in a semicircle). So

`sin/_C = sin/_ADB = {AB}/{AD} = c/d implies d = c/{sin/_C}`

Armed with this result, we first calculate the lengths of the three sides of the given triangle. Let us label the vertices `(3,7)`, `(2,9)`, and `(8,4)` by A, B, and C, respectively, Then

`a^2 = (2-8)^2+(9-4)^2 = 36+25 = 61 implies a = sqrt{61}`
`b^2 = (3-8)^2+(7-4)^2 = 25 + 9= 34 quad implies b = sqrt{34}`
`c^2 = (3-2)^2+(7-9)^2 = 1+4 = 5 qquad quad implies c = sqrt{5}`

We also need one of the angles, say `/_A`. For this we appeal to the cosine law for triangles :

`a^2 = b^2+c^2-2bc cos A implies cos A = {b^2+c^2-a^2}/{2bc}`

With our numbers, we have

`cos A = {34+5-61}/{2sqrt34sqrt5} = -11/sqrt170`

Finally

`d^2 = a^2/{sin^2A} = a^2/(1-cos^2A)= 61/(1-121/170)=61/49 times 170`

So, the area of the circumscribed circle is

`A = pi d^2/4=pi/4 times 61/49 times 170 ~~ 166.2`