Expected area of circle defined by 3 random vertices in unit square?

Question

I was looking at expected area of a triangle determined by randomly placed points ... and I was wondering if when picking 3 random points in the unit square, instead of drawing the triangle who's vertices were the 3 points, you drew the circle that past through the 3 points. What is the expected area of the circle, and what is the expected area of overlap between the circle and the square the points are chosen in?

Answer 1

The expected area is infinite.

Wikipedia shows how to find the circumcircle equation from the Cartesian coordinates of the vertices of the triangle. To simplify the calculation, let's fix two points at $A=(0,0)$ and $B=(0,1)$ and see whether the expected area of a triangle with a third point $C=(C_x,C_y)$ in the unit square is finite. If it isn't, the same type of singularity will also appear if the first two points are picked randomly.

For this case, the equations for the centre of the circumcircle reduce to

\begin{eqnarray} v_x&=&\frac12\;,\\ v_y&=&\frac{C_x^2+C_y^2-C_x}{C_y}\;. \end{eqnarray}

The area of the circumcircle is $\pi|C|^2$ and thus diverges as $C_y^{-2}$. If $C_y$ is chosen uniformly in the unit square, the corresponding integral diverges.