As to your question about the shape of a chain draped over two pulleys, I suggest that you look first at the much simpler problem of determining the shape of a chain of length L suspended from two points within a vertical plane. Here is how it is done.

Let (x1,y1) and (x2,y2) be the Cartesian coordinates of the suspension points, and let y = f(x) be the equation of the chain.

The curve goes through the points (x1,y1) and (x2,y2). That gives us two equations:

eq1 := y1 = f(x1),

eq2 := y2 = f(x2).

The curve's length is calculated by the usual curve length formula from calculus:

eq3 := int( sqrt(1 + f'(x)^2), x=x1..x2) = L;

Okay, so far we have specified the chain's geometry. Now the physics. It's possible to show, but I will not go through it here, that the chain's static equilibrium is expressed through the differential equation f''(x) = k*sqrt(1+f'(x)^2), where k is to be determined. In the worksheet below we see that this differential equation's general solution is a hyperbolic cosine.

Solving the differential equation brings along two integration constants c1 and c2. So altogether, the differential equation's solution involves three unknown c1, c2, k. We solve the system of equations {eq1, eq2, eq3} to determine those three unknowns.

The system is transcendental and has no useful symbolic solution, but we may solve the system numerically for any desired set of parameters (x1,y1), (x2,y2), and L. See the worksheet below.

Going back to your question on the shape of a chain draped over two pulleys, that can be done in a similar way. The curve will be a catenary since that's the solution of the differential equation of the chain's equilibrium. I haven't done the calculations myself but the idea would be that instead of passing the curve from the points (x1,y1), (x2,y2), we need to find the curve which is tangent to the circles c1 and c2 that represent the pulleys. The solution should be within reach but not worth the trouble unless you really need it for some purpose.

Finally, regarding your question whether the points labeled A in your diagram in your worksheet are on horizontal diameters, the answer is no. Just consider a chain which is tightly wrapped around the pulleys. The points A will be on vertical diameters in that case.

restart;
# The differential equation of a hanging chain
de := diff(y(x),x,x) = k*sqrt(1 + diff(y(x),x)^2);
# Solving the differential equation we obtain the equation of the hanging
# chain which involves three parameters, _C1, _C2, k:
dsolve(de):
f := unapply(rhs(%), x);
# We want the chain to go through the points (x1,y1) and (x2,y2):
eq1 := y1 = f(x1);
eq2 := y2 = f(x2);
# Additionally, we want the chain's length be a specified amount, L.
# The chain's length is calculated by the curve length formula from calculus:
int(sqrt(1+D(f)(x)^2), x):
simplify(%) assuming real:
eq3 := eval(%, x=x2) - eval(%, x=x1) = L;
# Now we need to solve the system of three equations {eq1, eq2, eq3} for the
# three unknowns _C1, _C2, k. There is no symbolic solution, so we do a
# numerical illustration.
params := { x1=0, y1=0, x2=10, y2=2, L=15 };
sys := eval({eq1,eq2,eq3}, params);
sol := fsolve(sys, {_C1,_C2,k}, {k=0..infinity});
#
# And now plot the solution
%plot(f(x), x=x1..x2, scaling=constrained):
eval(%, sol union params):
value(%);

Download worksheet: mw.mw