the problem 9.
(I take Radio 1 to make easier the problem)
first start with 2 circles one of Radius 1 center (0;0) and the other of radius R center (0;1) and define 3 points and 2 functions like picture
F(x) is the top part of the small circle, ##y= \sqrt{1-{x}^{2}}##
h(x) is the bottom part of the circle of radius r, ##y=- \sqrt{{R}^{2}-{x}^{2}}+1##
F is the center of the big circle (0;1)
G is the point 1-R in y axys
I is the point whit 2 circles intersected, h(x)=F(x) or $$- \sqrt{{R}^{2}-{x}^{2}}+1= \sqrt{-{x}^{2}+1}$$
$$({R}^{2}-{x}^{2})-2\, \sqrt{{R}^{2}-{x}^{2}}+1=-{x}^{2}+1$$
$${R}^{2}-{x}^{2}=1/4\,{R}^{4}$$
$$x=1/2\,R \sqrt{-{R}^{2}+4}$$
so I is the point ##1/2\,R \sqrt{-{R}^{2}+4}## in x axys
now we can see that R > 1 because otherwise would have less area than half of the circle
and R<sqrt(2) because otherwise would have more area than half of the circle.
we know the area of the small circle, is pi*1^2 = pi. so the area between the 2 circles must be pi/2, and the area FGI must be pi/4
now the hard part: $$\int_{0}^I \int_{h(x)}^{f(x)} \, dy \, dx= \pi/4$$
$$\int_{0}^{1/2\,R \sqrt{-{R}^{2}+4}} \int_{- \sqrt{{R}^{2}-{x}^{2}}+1}^{ \sqrt{-{x}^{2}+1}} \, dy \, dx= \pi/4$$
$$\int_{0}^{1/2\,R \sqrt{-{R}^{2}+4}}\! \sqrt{-{x}^{2}+1}+ \sqrt{{R}^{2}
-{x}^{2}}-1\,{\rm d}x= \pi/4$$
now I have to admit that I received some help to solve this integral. so will put the result assuming that 1 < R < sqrt(2)
and we Finally get: $$-1/4\,R\sqrt {-{R}^{2}+4}+1/2\,\arcsin \left( 1/2\,R\sqrt {-{R}^{2}+4}
\right) +1/2\,{R}^{2}\arcsin \left( 1/2\,\sqrt {-{R}^{2}+4} \right)= \pi/4$$
and this ecuation have a solution of R=1.158728473...
so the proportion is R=1.158728473... of the field radius.I really liked not having to ask for help to solve that integral and the numerical solution for the ecuation.
but as it was not the main problem decided not to give much importance,
I would like to see a solution that did not have calculus, and if that 1.158728473... have a exact form.
thanks =D