A2=πr2, A1/A2=0.5=½ (given), so A1=½πr2. Let p=r/R, the ratio to be found, then r=pR and A1=½πp2R2.
Let the centre of the larger circle be O1 and that of the smaller circle O2. Let the line segment O1O2=a (the separation of the centres of the circles). Let the circles intersect at common points A and B on both circles. O1A=O1B=R, O2A=O2B=r=pR. Join A to B, then chord AB, common to both circles, is perpendicular to O1O2 and AB splits the area A1 into two parts A1R and A1L, where L and R indicate left and right parts of A1. Let AB subtend angle θ1 in the larger circle and θ2 in the smaller circle, then a=Rcos(½θ1)+rcos(½θ2).
The area A1 can be expressed as the sum of the differences in area between a circular segment and an isosceles triangle in each circle.
For the larger circle: A1R=½(R2θ1-R2sin(θ1))=½R2(θ1-sin(θ1)) (that is, segment area minus triangular area)
For the smaller circle: A1L=½(r2θ2-r2sin(θ2))=½p2R2(θ2-sin(θ2)).
So A1=½πp2R2=½R2(θ1-sin(θ1))+½p2R2(θ2-sin(θ2))=½R2(θ1-sin(θ1)+p2(θ2-sin(θ2))),
πp2=θ1-sin(θ1)+p2(θ2-sin(θ2)), π=(θ1-sin(θ1))/p2+θ2-sin(θ2), by dividing through by ½R2 and then by p2.
O1O2 bisects θ1 and θ2. AB is a common chord and also the common base of two isosceles triangles.
AB=2Rsin(θ1/2)=2rsin(θ2/2), p=r/R=sin(θ1/2)/sin(θ2/2), from which it's clear that, since R>r (p<1), θ2>θ1.
p2=sin2(θ1/2)/sin2(θ2/2), or, using the identity cos(2A)=1-2sin2(A), that is, sin2(A)=½(1-cos(2A)), and substituting A=½θ1 and A=½θ2:
p2=(1-cos(θ1))/(1-cos(θ2)). Therefore:
π=(θ1-sin(θ1))(1-cos(θ2))/(1-cos(θ1))+θ2-sin(θ2).
This equation contains two variables so there is no unique solution for θ1 and θ2, implying that there is no unique solution for r/R.
EXAMPLE
The picture below illustrates the example.
Let θ1=1.7980 radians (103° approx) and θ2=0.8π (144°), then p=r/R=0.823 approx. There are many solutions: if the small circle is reduced in size without reducing the large circle, then it's easy to see that, by adjusting the distance between the centres of the circles, the large circle can be made to cover exactly half the area of the small circle. This changes the sizes of the angles in the triangles and r/R also changes, hence many solutions.