Looks Like Part Cardioid!

There is probably a faster solution than this one.

Let us define the moving point as $(t,0)$ , where we change $t$ as we choose. The shortest distance between the moving point and the bottom half of the circle can be found by minimizing the function:

$d(x)=(x-t)^{2}+(-\sqrt{1-x^{2}}+1)^{2}$

This can be done by taking the derivative and setting it equal to 0.

$\begin{aligned} 0&=2(x-t)+2(-\sqrt{1-x^{2}}+1)\cdot(\frac{x}{\sqrt{1-x^{2}}})\\ 0&=x-t-\frac{x\sqrt{1-x^{2}}-x}{\sqrt{1-x^{2}}}\\ 0&=x-t-x+\frac{x}{\sqrt{1-x^{2}}}\\ t&=\frac{x}{\sqrt{1-x^{2}}}\\ t^{2}&=\frac{x^{2}}{1-x^{2}}\\ x^{2}&=t^{2}-x^{2}t^{2}\\ t^2&=x^{2}(1+t^2)\\ x&=\sqrt{\frac{t^{2}}{1+t^{2}}}\\ \sqrt{d(\sqrt{\frac{t^{2}}{1+t^{2}}})}&=\sqrt{t^{2}+1}-1\\ \end{aligned}$

This is the distance from the moving point to the circle. If we make another circle $(x-t)^{2}+y^{2}=(\sqrt{t^{2}+1}-1)^2$ ), we see that this and the given circle intersect at one point. This new circle will be called circle 1.

$\begin{cases} y=-\sqrt{1-x^{2}}+1\\ y=\sqrt{(\sqrt{t^{2}+1}-1)^{2}-(x-t)^{2}}\\ \end{cases}$

$\begin{aligned} -\sqrt{1-x^{2}}+1&=\sqrt{(\sqrt{t^{2}+1}-1)^2-(x-t)^{2}}\\ 2-x^{2}-2\sqrt{1-x^2}&=(\sqrt{t^{2}+1}-1)^2-(x-t)^{2}\\ 2-x^{2}-2\sqrt{1-x^2}&=t^{2}+2-2\sqrt{t^{2}+1}-x^{2}+2xt-t^{2}\\ -2\sqrt{1-x^{2}}&=-2\sqrt{t^{2}+1}+2xt\\ \sqrt{1-x^{2}}&=\sqrt{t^{2}+1}-xt\\ 1-x^{2}&=t^{2}+1-2xt\sqrt{t^{2}+1}+x^{2}t^{2}\\ 0&=x^{2}(t^{2}+1)-x\cdot 2t\sqrt{t^{2}+1}+t^{2}\\ x&=\frac{t\sqrt{t^{2}+1}}{t^{2}+1}\\ \end{aligned}$

Plugging in this $x$ value into the equation of the bottom half of the given circle, we get that the point of intersection of the two circles is $(\frac{t\sqrt{t^{2}+1}}{t^{2}+1},-\frac{1}{\sqrt{1+t^{2}}}+1)$ .

The mirror image of the moving point will be on the circle $(x-t)^{2}+y^{2}=4(\sqrt{t^{2}+1}-1)^{2}$ , as it has twice the radius of circle 1. We will call this circle 2.

The line through the center of circle 2 and the point of intersection we found earlier is $y=1-\frac{x}{t}$ .

Now we need to find the intersection of circle 2 and this line. After solving this system of equations in a similar way as before, we get that the line and circle 2 intersect at $(\frac{2t\sqrt{t^2+1}}{t^2+1}-t,2-\frac{2\sqrt{t^2+1}}{t^2+1})$ . This is exactly the parametric form of the ribbon shaped curve.

The curve has a vertical tangent when $t=\sqrt{2^{\frac{2}{3}}-1}$ . The curve intersects the $y$ -axis at $t=0$ and $t=\sqrt{3}$ .

If we say that $f(t)=\frac{2t\sqrt{t^2+1}}{t^2+1}-t$ and $g(t)=2-\frac{2\sqrt{t^2+1}}{t^2+1}$ , the area of the loop is

$\begin{aligned} &2(\int_{\sqrt{3}}^{\sqrt{2^{\frac{2}{3}}-1}}g(t)f'(t)dt-\int_{0}^{\sqrt{2^{\frac{2}{3}}-1}}g(t)f'(t)dt)\\ &=\boxed{0.653} \end{aligned}$