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?
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@InterstellarProbe Isn't the probability of that event occurring zero? – TheHolyJoker Jan 17 '20 at 20:01
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2@TheHolyJoker I thought the OP was choosing points on the boundary of the unit square. You are correct. My mistake. However, circles are a bit different to calculate than triangles. There may be a division by zero issue to consider. As a circle approaches three collinear points, wouldn't the area approach infinite? This is very different from a triangle. – SlipEternal Jan 17 '20 at 20:13
1 Answers
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.

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What about the average area of overlap between the circle and square? – PlatinumStorm Jan 19 '20 at 19:11
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That's of course finite, but I think it will be tough to compute, with a lot of casework. If you're really interested in a closed form for that, I think you should ask that as a new question. – joriki Jan 19 '20 at 20:57