I got this question from mind your decision channel in youtube , and there the tutor has provided an elegant solution with euclidean geometry.
I am trying to work it out with coordinate geometry .
The figure is fairly symmetric and we need ratio of radius.
Taking the vertex of quadrat as origin, we write equation of bigger circle as $$x^2+y^2=R^2$$
Taking the center of the smaller semicircle as (a,a), the equation of the circle touching both the axes is $$x^2+y^2-2ax-2ay+a^2=0. $$
Now ,distance between center should be equal to
$a\sqrt2=R-a-$sagitta of the larger circle
or $$a\sqrt2=R-a-((R-\sqrt{R^2-a^2})$$
but this is not fetching the correct result.
can you point out my mistake ?
Thanks.
I don't see why $a\sqrt{2}$, the distance from the quarter-circle's right-angle to the red semicircle's base's centre, should be equal to $R-a$, i.e. the quarter-circle with the semicircle's radius subtracted. That's equivalent to claiming the larger circle's radius joining the two centres has an excess length $a$ beyond the semicircle's base, which doesn't look true.
What you should do instead is note that the shapes' equations imply $x+y=\frac{a^2+R^2}{2a}$ is the equation of the semicircle's base, and the endpoints satisfy $xy=\frac{(x+y)^2-x^2-y^2}{2}=\frac{(a^2-R^2)^2}{8a^2}$. Thus $x,\,y$ are, in some order, the roots of $t^2-\frac{a^2+R^2}{2a}t+\frac{(a^2-R^2)^2}{8a^2}=0$ (changing the order reflects one endpoint in $y=x$ to give the other). These roots are $$t_\pm:=\frac{a^2+R^2\pm\sqrt{6a^2R^2-a^4-R^4}}{4a}.$$The squared distance between $(t_+,\,t_-)$ and $(t_-,\,t_+)$ is $$4a^2=2(t_+-t_-)^2=\frac{6a^2R^2-a^4-R^4}{2a^2}.$$This rearranges to $0=(3a^2-R^2)^2$, i.e. $a=R/\sqrt{3}$.
Now we've proven that, let's answer the original question: $$\frac{\frac{\pi}{2}\left(\frac{R}{\sqrt{3}}\right)^2}{\frac{\pi}{4}R^2}=\frac{2}{3}.$$