Question

A triangle has vertices `A(a,b )`, `C(c ,d )`, and `O(0 ,0 )`. What is the equation and area of the triangle's circumscribed circle?

Answer 1

`(x-p)^2 + (y-q)^2 = s quad` where

`p = {d (a^2+b^2) - b(c^2+d^2)}/{2(ad-bc) }`

`q = {a(c^2+d^2)-c(a^2+b^2)}/{2(ad-bc) }`

`s = ((a^2 + b^2) (c^2 + d^2) ( (a-c)^2 + (b-d)^2 ))/(4 (ad-b c)^2)`

`A = pi s`

Explanation:

I generalized the question; let's see how that goes. I left one vertex at the origin, which makes it a bit less messy, and an arbitrary triangle is easily translated.

The triangle is of course totally inessential to this problem. The circumscribed circle is the circle through the three points, which happen to be the three vertices. The triangle does make a surprise appearance in the solution.

Some terminology: the circumscribed circle is called the triangle's circumcircle and its center the triangle's circumcenter .

The general equation for a circle with center `(p,q)` and squared radius `s` is

`(x-p)^2 + (y-q)^2 = s`

and the area of the circle is `A = pi s.`

We have three unknowns `p,q,s` and we know three points, so we get three equations:

`p^2+q^2=s quad` because the origin is on the circle.

`(a-p)^2 + (b-q)^2 = s`

`(c-p)^2 + (d-q)^2=s`

Let's solve the simultaneous equations. Let's turn them into two linear equations by expanding and subtracting pairs, which amounts to losing `p^2+q^2` on the left and `s` on the right.

`a^2 - 2ap + p^2 + b^2 - 2aq + q^2 = s`

Subtracting,

`a^2 + b^2 - 2ap - 2bq = 0`

`1/2 (a^2+b^2) = ap +bq`

Similarly,

`1/2( c^2 + d^2) = cp + dq`

That's two equations in two unknowns. `AX=K` has solution `X=A^{-1} K.` I remember the two by two matrix inverse which I don't know how to format,

`A^{-1} = 1/{ad-bc} ( stackrel{ d, -b} { -c, a})`

For us that means

`p = {d (a^2+b^2) - b(c^2+d^2)}/{2(ad-bc) }`

`q = {a(c^2+d^2)-c(a^2+b^2)}/{2(ad-bc) }`

and a squared radius of

`s = p^2 + q^2`

`s = { (d (a^2+b^2) - b(c^2+d^2) )^2 + ( a(c^2+d^2)-c(a^2+b^2))^2 }/{ 4 (ad-bc)^2}`

`s = ((a^2 + b^2) (c^2 + d^2) ( (a-c)^2 + (b-d)^2 ))/(4 (ad-b c)^2)`

so an area of `pi` times that amount.

We can see the expression become more symmetrical if we consider what happens for the arbitrary triangle `(A,B),(C,D),(E,F).` We set `a=A-E,` `b=B-F,` `c=C-E,` `d=D-F` but I won't work that out now.

I'll note the numerator of `s` is the product of the three squared lengths of the triangle's sides, and the denominator of `s` is sixteen times the squared area of the triangle.

In Rational Trigonometry squared lengths are called quadrances and sixteen times the squared area is called the quadrea. We found the quadrance of the radius of the circumcircle is the product of the quadrances of the triangle divided by its quadrea.

If we just need the radius or area of the circumcircle, we can summarize the result here as:

The squared radius of the circumcircle is the product of the squared lengths of the triangle divided by sixteen times the triangle's squared area.

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