Almost equal

Due to symmetry, we only need to consider the firs half quadrant of the graph, where the ratio $\dfrac {A_{\color{#20A900}\text{green}}}{A_{\color{#3D99F6}\text{blue}}}$ is the same.

We know that the point $P$ , where the circle intersects the hyperbola, satisfies the system of equations below.

$\begin{cases} \text{Hyperbola:} & x^2 - y^2 = \dfrac {r^2}4 & ...(1) \\ \text{Circle:} & x^2 + y^2 = r^2 & ...(2) \end{cases}$

From $(1)+(2): \ 2x^2 = \dfrac 54 r^2$ $\implies x = \sqrt{\frac 58}r$ . Since $x^2 + y^2 = r^2$ , $\implies y = \sqrt{\frac 38} r$ and $P \left(\sqrt{\frac 58}r, \sqrt{\frac 38}r \right)$ .

Then the green area is given by:

$\begin{aligned} A_{\color{#20A900}\text{green}} & = \int_\frac r2^{\sqrt{\frac 58}r} \sqrt{x^2 - \frac {r^2}4}\ dx + \int_{\sqrt{\frac 58}r}^r \sqrt{r^2 - x^2}\ dx \\ & = {\color{#3D99F6}\frac r2 \int_\frac r2^{\sqrt{\frac 58}r} \sqrt{\left(\frac {2x}r\right)^2 - 1}\ dx} + \color{#D61F06} r \int_{\sqrt{\frac 58}r}^r \sqrt{1 - \left(\frac xr\right)^2}\ dx & \small \color{#3D99F6} \text{Let } \frac {2x}r = \sec \theta \implies dx = \frac r2 \sec \theta \tan \theta\ d\theta \\ & = {\color{#3D99F6}\frac {r^2}4 \int_0^{\tan^{-1}\sqrt{\frac 32}} \sec \theta \tan^2 \theta\ d\theta} + \color{#D61F06} r^2 \int_{\tan^{-1} \sqrt{\frac 53}}^\frac \pi 2 \cos^2 \phi \ d\phi & \small \color{#D61F06} \text{Let } \frac xr = \sin \phi \implies dx = r \cos \phi\ d\phi \\ & = {\color{#3D99F6}\frac {r^2}4 \sec \theta \tan \theta\ \bigg|_0^{\tan^{-1}\sqrt{\frac 32}} - \frac {r^2}4 \int_0^{\tan^{-1}\sqrt{\frac 32}} \sec^3 \theta\ d\theta} + \frac {r^2}2 \int_{\tan^{-1} \sqrt{\frac 53}}^\frac \pi 2 (1+ \cos 2\phi)\ d\phi & \small \color{#3D99F6} \text{By integration by parts} \\ & = \frac {\sqrt{15}}8r^2 - {\color{#3D99F6} \frac {r^2}4 \left[\frac {\sin \theta \sec^2 \theta}2 + \frac 12 \int \sec \theta\ d\theta \right]_0^{\tan^{-1}\sqrt{\frac 32}}} + \frac {r^2}2 \left[ \phi + \frac 12 \sin 2\phi \right]^\frac \pi 2_{\tan^{-1} \sqrt{\frac 53}} & \small \color{#3D99F6} \text{By reduction formula} \\ & = \frac {\sqrt{15}}8r^2 - \frac {\sqrt{15}}{16}r^2 - \frac {r^2}8 \ln (\tan \theta + \sec \theta) \bigg|_0^{\tan^{-1}\sqrt{\frac 32}} + \frac \pi 4 r^2 - \frac {r^2}2 \tan^{-1} \sqrt{\frac 53} - \frac {\sqrt{15}}{16}r^2 \\ & = \frac {r^2}{16} \left(4\pi - \ln \left(4+\sqrt{15}\right) - 8 \tan^{-1} \sqrt{\frac 53} \right) \\ & \approx 0.200564201r^2 \end{aligned}$

Then $\dfrac {A_{\color{#20A900}\text{green}}}{A_{\color{#3D99F6}\text{blue}}} = \dfrac {A_{\color{#20A900}\text{green}}}{\frac \pi 8 r^2 - A_{\color{#20A900}\text{green}}} \approx \dfrac {0.200564201r^2}{0.392699082r^2 -0.200564201r^2 } \approx \boxed{1.044}$

The green area is four times the value of the integral:

$\displaystyle \int_{-\frac{\sqrt{6}r}{4}}^{\frac{\sqrt{6}r}{4}}[f(x)-g(x)]\,dx=2 \times \int_{0}^{\frac{\sqrt{6}r}{4}}\left(\sqrt{r^2-x^2}-\sqrt{x^2+\left(\frac{r}{2}\right)^2}\right)\,dx= \\ =2 \times \left[\dfrac{1}{2} x \sqrt{r^2-x^2}+r^2\arcsin{\frac{x}{r}}- \dfrac{1}{2} x \sqrt{x^2+\left(\frac{r}{2}\right)^2}-\left(\frac{r}{2}\right)^2\sinh^{-1}{\frac{2x}{r}} \right]_{0}^{\frac{\sqrt{6}r}{4}}=0.40113r^2 \\ \implies \text{green area}=4\times 0.40113r^2=1.60451r^2$

$\text{blue area}=\pi r^2-\text{green area}=1.53708r^2$

The desired ratio is $\dfrac{\text{green area}}{\text{blue area}}=\dfrac{1.60451r^2}{1.53708r^2}=\boxed{1.04387}$

This is a replicated rectangular hyperbola. As noted elsewhere, it is sufficient to work in the lower octant of the first quadrant and that any radius could be chosen. I used a radius of 1. By working in polar coordinates, the green area can be computed in one integral, with the hyperbola being at $r=\frac{1}{2} \sqrt{\sec (2 \theta )}$ , the circle being at $r=1$ and $0\leq \theta \leq \frac{1}{2} \cos ^{-1}\left(\frac{1}{4}\right)$ , which is the angle at which the hyperbola meets the circle. $\text{green}=8 \int _0^{\frac{1}{2} \cos ^{-1}\left(\frac{1}{4}\right)}\int _{\frac{1}{2} \sqrt{\sec (2 \theta )}}^1\,r\,dr\,d\theta\Rightarrow 2 \arccos{\frac14}-$ $\text{arctanh}{\sqrt{\frac35}}$ . The answer: $\frac{\text{green}}{\pi -\text{green}} \approx 1.04387189088098$ .

I thought almost equal area implies their quotient to be close to 1. Then i thought circle implies pi. Pi/3=1.0472 - correct.