Let the center to center distance be ##d## and ##r## the smaller circle radius and ##R## the larger circle radius so ##d+r=R##. We know ##d^2=r^2+r^2## or ##d=\sqrt{2}r##. Solving for ##r## in terms of ##R## we get $$r=\frac{R}{\sqrt{2}+1}$$ therefore the fraction that is shaded is $$\frac{\pi r^2}{\frac{1}{4}\pi R^2}= 4(\frac{r}{R})^2= \frac{4}{(\sqrt{2}+1)^2}\approx 0.686$$
In the 3D case, ##d=\sqrt{3}r## giving $$r=\frac{R}{\sqrt{3}+1}$$ now we have the fraction of volume of a sphere to an octant which is $$\frac{\frac{4}{3} \pi r^3}{\frac{1}{8}\frac{4}{3}\pi R^3}= 8(\frac{r}{R})^3= \frac{8}{(\sqrt{3}+1)^3}\approx0.392$$
In the case of 4D, ##d=\sqrt{4}r=2r## then ##r=\frac{R}{3}##. In the general case ##d=\sqrt{n}r## and the ratio $$\frac{r}{R}=\frac{1}{\sqrt{n}+1}$$
The general equation for hypersphere volume is
$$V_n(R) = \frac{\pi^{n/2}}{\Gamma\!\left(\frac{n}{2} + 1\right)} \, R^n$$
Note that in higher dimensions the number of orthants (quadrants in 2D and octant in 3D) goes as ##2^n##
Thus the fraction of the hypersphere of dimension ##n## to its orthant is $$\frac{2^n}{(\sqrt{n}+1)^n}$$
As ##n## gets larger, this fraction tends towards zero as does the ratio of ##r## to ##R## and a hypersphere volume in general.
I think that should be ##\approx##Why?
I don't see how this diagram helps visualize how the smaller sphere touches the face of the octant.
Regarding ##d=\sqrt{3}r##, the vector from the origin to the center of the small sphere has x,y,z components each equal to ##r##.
Regarding the graph, it just visualizes the smaller size.
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