Locate origins of three circles with given radii and intersection relations, all three lying within a circle of area/probability 1 Say I have a circle D of area/probability 1 (radius $\frac{1}{\sqrt{\pi}}$) centered at (0,0). I also have a circle A of area $\frac{8 \pi}{27 \sqrt{3}} \approx 0.537422$ and two circles B and C, each of area $\frac{1}{6} \approx 0.16667$, which I want to position entirely within D, so that the three circles satisfy certain intersection relations. The area of intersection of $A$ with $B$, and also $A$ with $C$ is small, that is $\frac{4}{9} + \frac{4 \pi}{27 \sqrt{3}} +\frac{\log{3}}{6} \approx 0.0073686$, with these two small areas not intersecting at all. Further, B and C have intersection $\frac{1}{9} \approx 0.11111$ and union $\frac{2}{9} \approx 0.22222$. So, what choice(s) of origins for A, B and C--if any--will give such areas/probabilities? Would further information--which I have--on the various intersection/union relations be helpful/needed?
The probabilities above have quantum-information-theoretic interpretations. This is expanded upon in essentially the same question (with same circle notation A, B, C, D) posed two months ago in a graphics/Venn diagram framework
https://mathematica.stackexchange.com/questions/198019/create-a-venn-diagram-showing-the-relations-of-three-sets-of-quantum-states
but not yet fully successfully addressed.
Per the request of Moti, here are a couple of attempts
https://www.wolframcloud.com/obj/4abd389b-9492-446d-8842-8a9fcfb16cfb
The circles are of the correct area, but the intersection of the two smaller circles (B, C) within A should be 0. (I have an objective function based on a weighted sum of the pairwise overlaps of the three circles which I'm trying to minimize through a random-number-based procedure. So, it would be preferable to have an exact procedure, if possible. I'm not sure how to incorporate the [three circle] zero-intersection requirement.)
 A: This is more of a general guideline to solve the problem than an actual solution. :-(
First, it should be clear that the common area inclosed by 2 circles of given, fixed radii $r,R$ (assume $r \le R$,w.l.o.g.) it completely determined by the distance $d$ of their midpoints. Given those values, the area common between the circles can be found and numerically calculated, you can see the result and how it is derived at Wolfram.
What's not clear from the formula given there, is that the common area is a strictly monotonically decreasing function for
$$R-r \le d \le R+r.$$
Outside this interval, the situation is clear: If $d > R+r$, then the circles don't intersect at all, so the common area is $0$. If $d < R-r$, the smaller circle is completely inside the bigger one, the the common area is the smaller circle's area ($\pi r^2)$.
Inside the interval it is clear that the common area goes from $\pi r^2$ on the left end of the interval to $0$ on the right hand. An argument can be made (that I'm not giving here, that I'm 95% sure of, without having written it formally down), that inside the interval the function is strictly deacreasing.
That means your knowledge on the exact value of the common area of the intersection of 2 circles with known radii determines the distance of their midpoints exactly. If we call the midpoints of the circles $A,B,C$ $M_A,M_B$ and $M_C$, resp., you can now (numerically) calculate this distance between any two of them by the information you have on their radii and their common area.
Of course, the 3 distances between $M_A,M_B$ and $M_C$ determine their position in the plane up to congruence, so the postions of the 3 circles $A,B$ and $C$ is also determined uniquly up to congruence.
That means if the 3 circles $A,B$ and $C$ have a common interior point, which would mean positive area of intersection, is already completely determined by the information given, you don't have any 'wiggle room' to arrange them.
I guess that the picture you gave has already been made such that the 3 areas of intersection are correct. You can see that the 2 postions (by visual inspection) aren't 'different', they are just rotatated inside the big circle $D$. If that's true, it would mean there is no solution to your problem.
A: This is by no means a direct answer to the question posed. It is a "work-around"--introducing new details--of the main presentational difficulty, which seemed to be the depiction of the void intersection of the circles B and C within the circle A. The values ("bound-entangled probabilities") $\frac{4}{9}+\frac{4 \pi }{27 \sqrt{3}}+\frac{\log (3)}{6} \approx 0.00736862$ of the intersections of A with B, and of A with C had been obtained through three-dimensional constrained (Boolean) integrations.
Issuing the Mathematica command
Show[RegionPlot3D[Q3>0&&Q1+3 Q2+2 Q3<1&&Q1^2+3 Q1 Q2+(3 Q2+Q3)^2<3 Q2+2 Q1 Q3&&Q1>3 Q2+4 Q3,{Q1,0,1/3},{Q2,0,1/3},{Q3,0,1/6 (2-Sqrt[3])},AxesLabel->{"Subscript[Q, 1]","Subscript[Q, 2]","Subscript[Q, 3]"}],RegionPlot3D[Q1>0&&Q2>0&&Q3>0&&Q1+3 Q2+2 Q3<1&&Q1^2+3 Q1 Q2+(3 Q2+Q3)^2<3 Q2+2 Q1 Q3&&1+2 Q3<2 Q1+3 Q2,{Q1,0,1/3},{Q2,0,1/3},{Q3,0,1/6 (2-Sqrt[3])},AxesLabel->{"Subscript[Q, 1]","Subscript[Q, 2]","Subscript[Q, 3]"}]]

where the constraints indicated were those employed in the integrations, gives us the output
https://www.wolframcloud.com/obj/e4094c7c-ecd9-4232-bff4-895506f2fe85
There are two objects in this plot--the volumes of each of which are equal to $\frac{1}{36}$ of $\frac{4}{9}+\frac{4 \pi }{27 \sqrt{3}}+\frac{\log (3)}{6} \approx 0.00736862$. The front one corresponds to the intersection of A with B, and the posterior one to A with C. So, as desired, the non-intersecting character of the two objects is revealed. 
Perhaps the question as put was not the most effective way of proceeding in this subject matter context--per the comment of  Ingix: "Ultimately, I'm not sure exactly why you want that problem solved. If it is simply for 'explanatory purposes', a diagram will be enough, if it shows the correct relations, that is ∩∩=∅ ."
Additionally, here is a plot including also the "progenitor" of A, along with those of B and C.
https://www.wolframcloud.com/obj/34c06e57-f136-424f-852a-6c8ad5928b08
It can be seen that B and C contact A at different locations, so it can be readily concluded that ∩∩=∅.
