I have some limaçon with the polar equation $r=a+b\cos(\theta)$. I want to strecth the limaçon along some small vector $W=(c,d)$. By scretching along a small vector I mean the center of the coordinate system remains inside the limaçon. Now, the question is
this new closed curve still will be a limaçon? In any case, what would be the equation of this new closed curve?
I will appreciate any comment or solutions.
Given a curve $\gamma$ of parametric equations:
$$ \begin{cases} x = x(t) \\ y = y(t) \\ \end{cases} \; \; \; \; \; \; \text{for} \; t \in I \subseteq \mathbb{R} $$
in order for $\gamma$ to be scaled with respect to a point $C\left(x_C,\,y_C\right)$ by a factor $s_1 > 0$ in counterclockwise direction $\theta \in [0,\,2\pi)$ and by a factor $s_2 > 0$ in counterclockwise direction $\theta + \frac{\pi}{2}$, it's necessary:
translate $\gamma$ in such a way that $C$ coincides with the origin;
rotate $\gamma$ by $-\theta$ with respect to the origin;
scale $\gamma$ by a factor $s_1$ along the x-axis and by a factor $s_2$ along the y-axis with respect to the origin;
rotate $\gamma$ by $+\theta$ with respect to the origin;
translate $\gamma$ in such a way that $C$ returns to its initial position;
hence the curve $\gamma'$ of parametric equations:
$$ \small \begin{cases} x = x_C + \left(x(t) - x_C\right)\left(s_1\cos^2\theta + s_2\sin^2\theta\right) + \left(y(t) - y_C\right)\left(s_1 - s_2\right)\cos\theta\,\sin\theta \\ y = y_C + \left(y(t) - y_C\right)\left(s_1\sin^2\theta + s_2\cos^2\theta\right) + \left(x(t) - x_C\right)\left(s_1 - s_2\right)\cos\theta\,\sin\theta \\ \end{cases} \; \; \; \; \; \; \text{for} \; t \in I $$
that is, considering the versor $\hat{n}$ of components $n_1 = \cos\theta$ and $n_2 = \sin\theta$, we get:
$$ \small \begin{cases} x = x_C + \left(x(t) - x_C\right)\left(s_1\,n_1^2 + s_2\,n_2^2\right) + \left(y(t) - y_C\right)\left(s_1 - s_2\right)n_1\,n_2 \\ y = y_C + \left(y(t) - y_C\right)\left(s_1\,n_2^2 + s_2\,n_1^2\right) + \left(x(t) - x_C\right)\left(s_1 - s_2\right)n_1\,n_2 \\ \end{cases} \; \; \; \; \; \; \text{for} \; t \in I\,. $$
If $s_2 = 1/s_1$ there's a squeezing, i.e. the area enclosed by a closed curve is invariant.
In particular, considering a limaçon of parametric equations:
$$ \begin{cases} x(t) = \left(1 + 2\cos(t)\right)\cos(t) \\ y(t) = \left(1 + 2\cos(t)\right)\sin(t) \\ \end{cases} \; \; \; \; \; \; \text{for} \; t \in [0,\,2\pi) $$
whose centroid turns out to be $C\left(\frac{6}{5},\,0\right)$, if we want it to be scaled with respect to the centroid by a factor $s_1 = k$ in the anticlockwise direction $\frac{\pi}{3}$ and by a factor $s_2 = 1$ in the anticlockwise direction $\frac{5\pi}{6}$, we get:
$$ \begin{cases} x = \frac{3\,(1 - k)}{10} + \frac{k + 3}{4}\,x(t) + \frac{\sqrt{3}\,\left(k - 1\right)}{4}\,y(t) \\ y = \frac{3\sqrt{3}\,(1 - k)}{10} + \frac{3\,k + 1}{4}\,y(t) + \frac{\sqrt{3}\,\left(k - 1\right)}{4}\,x(t) \\ \end{cases} \; \; \; \; \; \; \text{for} \; t \in [0,\,2\pi) $$
where for $0 < k < 1$ a contraction is obtained, while for $k > 1$ a dilation is obtained.
Wanting to view everything through a gif, in Wolfram Mathematica 12.2 it's sufficient to write:
from which I would say that the purpose has been achieved! ^_^
About the last request in the comments, i.e. wanting to scale with respect to the centroid:
by a factor $1.5$ in the anticlockwise direction $\frac{\pi}{3}$;
by a factor $0.8$ in the anticlockwise direction $\frac{5\pi}{6}$;
by a factor $0.9$ in the anticlockwise direction $\frac{4\pi}{3}$;
it's necessary to divide the curve $\gamma$ into the parts placed above and below the line passing through the centroid $C_{\gamma}$ and of anticlockwise direction $\frac{5\pi}{6}$:
That's all.