I want to know if the equation I give here to model the transformation works as expected
For this post let $n$ be a positive integer greater than one
Define:
$$W=\frac{a}{2}\left(1-T\right)\cos\left(\left(n+1\right)b\right)+\frac{a}{2}\left(1-T\right)\cos\left(\left(n-1\right)b\right)+a\left(1-T\right)\cos b+T\sqrt{\frac{A_{1}}{\pi}}\cos b$$
$$Z=\frac{a}{2}\left(1-T\right)\left(n+1\right)\cos\left(\left(n+1\right)b\right)+\frac{a}{2}\left(1-T\right)\left(1-n\right)\cos\left(\left(1-n\right)b\right)+a\left(1-T\right)\cos b+T\sqrt{\frac{A_{1}}{\pi}}\cos b$$
$$B_{1}=A_{1}$$
$$B_{2}=A_{2}$$
$$q=0$$
$$A_{3}=\int_{0}^{2\pi}\left(\sqrt{\frac{B_{1}}{B_{2}}+q}W\right)\left(\sqrt{\frac{B_{1}}{B_{2}}+q}Z\right)db$$
$$X_{1}=\frac{a}{2}\cos\left(\left(n+1\right)u\right)+\frac{a}{2}\cos\left(\left(n-1\right)u\right)+a\cos u$$
$$Y_{1}=\frac{a}{2}\left(n+1\right)\cos\left(\left(n+1\right)u\right)+\frac{a}{2}\left(1-n\right)\cos\left(\left(1-n\right)u\right)+a\cos u$$
$$A_{1}=\int_{0}^{2\pi}X_{1}Y_{1}du$$
$$h\left(t\right)=f\left(t\right)+T\left(g\left(t\right)-f\left(t\right)\right)$$
$$X_{2}=\frac{a}{2}\left(1-T\right)\cos\left(\left(n+1\right)w\right)+\frac{a}{2}\left(1-T\right)\cos\left(\left(n-1\right)w\right)+a\left(1-T\right)\cos w+T\sqrt{\frac{A_{1}}{\pi}}\cos w$$
$$Y_{2}=\frac{a}{2}\left(1-T\right)\left(n+1\right)\cos\left(\left(n+1\right)w\right)+\frac{a}{2}\left(1-T\right)\left(1-n\right)\cos\left(\left(1-n\right)w\right)+a\left(1-T\right)\cos w+T\sqrt{\frac{A_{1}}{\pi}}\cos w$$
$$A_{2}=\int_{0}^{2\pi}X_{2}Y_{2}dw$$
Model the starting shape by $$f\left(t\right)=\left(\frac{a}{2}\cos\left(\left(n+1\right)t\right)+\frac{a}{2}\cos\left(\left(n-1\right)t\right)+a\cos t,\frac{a}{2}\sin\left(\left(n+1\right)t\right)+\frac{a}{2}\sin\left(\left(1-n\right)t\right)+a\sin t\right)$$
Model the resultant circle by $$g\left(t\right)=\sqrt{\frac{A_{1}}{\pi}}\left(\cos t,\sin t\right)$$
Then the continuous deformation of the starting shape into a circle centered at the origin can be modeled by the equation
$$H\left(t\right)=\sqrt{\frac{A_{1}}{A_{2}}}h\left(t\right)$$
where the $T$ in $h\left(t\right)=f\left(t\right)+T\left(g\left(t\right)-f\left(t\right)\right)$ goes from zero to 1.
For a given time slice at $T=T_0$ one obtains an intermediate curve whose enclosed area is the same as the enclosed area of the starting curve
This link to my desmos graph models the transformation using the equations above.
To verify that enclosed area is preserved, see that the $A_3$ remains constant as $T$ varies only if $q=0$