Circumcircle of degenerate triangle

Question

What is the correct way to think about a circumcircle of a degenerate triangle, i.e. one where all 3 vertices are collinear? Obviously whenever you choose 3 collinear points on the plane it's impossible to construct a circle around them, but from a theoretical point of view, could the degenerate triangle (specifically a line) be thought of as an infinitely small line (which would in turn degenerate into a point) lying on some hypothetical circle?

In that case, a circumcircle around a degenerate triangle like that would be a real projective line which would be an extension of the triangle; given the triangle is infinitely small, the circle would have to be relatively infinitely large and we'd never see any sort of curvature. Only when we'd "infinitely zoom out" would we see the triangle as a single point and the line as a circle. Is this problem even related to the real projective line?

Or is the case of a degenerate triangle just safely left undefined?

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Edit: I noticed some people have considered the case where two or more vertices of the triangle are equal, which was not what I meant but certainly not something I excluded in the question body. Nevertheless these examples led me to interesting results as well. Thanks for that!

Answer 1

A circle by three aligned points is a straight line, nothing else.

If you don't believe it, consider the pencil of circles by two fixed points and a moving one. When the moving point comes to alignment, the quadratic terms vanish from the equation. The curvature is zero.

E.g. The pencil by the points $(-1,0),(1,0),(0,t)$ has the equation

$$\begin{vmatrix}x^2+y^2&x&y&1\\1&-1&0&1\\1&1&0&1\\t^2&0&t&1\end{vmatrix}=0$$

or

$$t(x^2+y^2)+y(1-t^2)-t=0.$$

This is the equation of a circle of center $\left(0,\dfrac{t^2-1}{2t}\right)$ and radius $\dfrac{1+t^2}{2|t|}$, and it degenerates to the straight line $y=0$ when $t=0$.

Answer 2

The circumcircle of a degenerate triangle can be thought of intuitively as a circle which is infinitely large whose boundary consists of the infinite straight line which contains the degenerate triangle. Think about what happens (or use a graphics tool) to watch the size of the circumcircle as one of the angles gets closer and closer to 180 degrees. The circle gets bigger and bigger.

Answer 3

Sometimes it is convenient to consider the degenerate triangle with two coincided points as a degenerate isosceles triangle. The area of the isosceles triangle with sides $a,a,b$ is

\begin{align} S&=\tfrac14b\sqrt{4a^2-b^2} ,\\ \text{and }\quad R&=\frac{a^2b}{4\cdot\tfrac14b\sqrt{4a^2-b^2}} =\frac{a^2}{\sqrt{4a^2-b^2}} , \end{align}
and for the degenerate case when $b=0$, we naturally have

\begin{align} R&=\frac a2 \tag{1}\label{1} . \end{align}

Of course, as @Blue noted in the comments, there are infinitely-many circles and a line, passing through these three points, but \eqref{1} can be considered more appropriate in some practical cases, like for example, it is often convenient to consider that $0^0=1$.

But in other cases this would be a wrong choice for the value of circumradius.

Consider this construction:

$\triangle ABC$, $D,E\in AB$, $|CD|=u,\ |DE|=v,\ |CE|=w$, $\angle CDE=\angle CDB=\delta$. Let $R,R_1,R_2,R_3,R_4$ be the radii of circumscribed circles around $\triangle ABC,\ \triangle CAD,\ \triangle CDE,\ \triangle CEB$ and $\triangle CDB$, respectively. Then we have an identity

\begin{align} R_1\cdot R_3&= R\cdot R_2 \tag{2}\label{2} . \end{align}

Now, let's move the point $E$ toward $D$ until they coincide. Then $R$ and $R_1$ stay the same, $R_3$ becomes $R_4$. And what will happened to $R_2$? From this point of view, $R_2=\tfrac u2$ would be a wrong choice. More appropriate would be to consider this degenerate triangle $CDD$ as one with the fixed side length $u$ and fixed angle $\delta$:

\begin{align} S_{CDE}&=\tfrac12uv\sin\delta ,\\ R_{CDE}=R_2&= \frac{uvw}{4\cdot \tfrac12uv\sin\delta} =\frac{w}{2\sin\delta} \tag{3}\label{3} , \end{align}

so when $w\to u$, to get sensible result in \eqref{2}, we must have

\begin{align} R_{CDD}&= \frac{|CD|}{2\sin\delta} \tag{4}\label{4} , \end{align}

and \eqref{2} transforms into

\begin{align} R_1\cdot R_4&= R\cdot \frac{|CD|}{2\sin\delta} \tag{5}\label{5} . \end{align}

So, the definition for the circumradius of the degenerate triangles is indeed context-sensitive.