I want to know if the equation I provide actually does indeed model the transformation that I desire to model
Model the starting shape by $$f\left(t\right)=\left(a\left(\cos\left(2t\right)+1\right)\cos t,a\left(\cos\left(2t\right)+1\right)\sin t\right)$$
Model the resultant circle by $$g\left(t\right)=\sqrt{\frac{3}{2}}a\cdot\left(\cos\left(t\right),\sin\left(t\right)\right)$$
Define:
$$X=\frac{a}{2}\left(1-T\right)\cos\left(3u\right)+\left(\frac{3}{2}-\frac{3}{2}T+T\sqrt{\frac{3}{2}}\right)a\cos\left(u\right)$$
$$Y=\frac{3a}{2}\left(1-T\right)\cos\left(3u\right)+\left(\frac{1}{2}-\frac{T}{2}+T\sqrt{\frac{3}{2}}\right)a\cos\left(u\right)$$
$$A=\int_{0}^{2\pi}XYdu$$
$$h\left(t\right)=f\left(t\right)+T\left(g\left(t\right)-f\left(t\right)\right)$$
Then in order to ensure that the enclosed area remains constant throughout the transformation, model the transformation by
$$H\left(t\right)=\sqrt{\frac{3\pi}{2A}}ah\left(t\right)$$
Here I have a graph modeling the transformation using the above equations, where we can see the closed curve continuously deform as $T$ goes from zero to one.
As you may be able to guess, what I've done for the area calculation is apply Green's Theorem