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?
1 Answer
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
and the area of the circle is
We have three unknowns
Let's solve the simultaneous equations. Let's turn them into two linear equations by expanding and subtracting pairs, which amounts to losing
Subtracting,
Similarly,
That's two equations in two unknowns.
For us that means
and a squared radius of
so an area of
We can see the expression become more symmetrical if we consider what happens for the arbitrary triangle
I'll note the numerator of
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.