Equation of a stretched-squeezed limaçon! I have a limaçon with the polar equation $r=a+b\cos(\theta)$. I want to strecth it in the direction of the vector $W=(c,d)$ and squeeze it in the opposite direction of $W$. I have a figure here which may help. The black one is the origin limaçon and the green one is the squeezed and sctetched one. Here $a=3.2$ and $b=2$.

There is a solution for the case that I squeeze or scretch in the direction of $W$.
Now, the question is:

what would be the equation of this new closed curve? will this new closed curve still will be a limaçon?

I will appreciate any comment or solutions.
 A: tl;dr: No, a stretched limaçon is not a limaçon.

Let's be a bit more "generous" and call a limaçon any polar graph
$$
r = a + b\cos(t + t_{0}),
$$
i.e., that can be obtained by rotating the polar graph $r = a + b\cos t$ about the origin by $t_{0}$. A general limaçon therefore has parametric form
\begin{align*}
  x(t) &= (a + b\cos(t + t_{0}))\cos t, \\
  y(t) &= (a + b\cos(t + t_{0}))\sin t.
\end{align*}
It appears stretching amounts to applying a specific "orthogonally diagonalizable" linear transformation. Again, let's be more flexible and assume we're allowed to apply an arbitrary invertible linear transformation
$$
\left[\begin{array}{@{}c@{}}
    u \\
    v \\
  \end{array}\right]
= \left[\begin{array}{@{}cc@{}}
    A & B \\
    C & D \\
  \end{array}\right]
\left[\begin{array}{@{}c@{}}
    x \\
    y \\
  \end{array}\right]
= \left[\begin{array}{@{}c@{}}
    Ax + By \\
    Cx + Dy \\
  \end{array}\right],\quad
AD - BC \neq 0.
$$
The parametric equations for a "stretched" limaçon are therefore
\begin{align*}
  u(t) &= (a + b\cos t)(A\cos t + B\sin t), \\
  v(t) &= (a + b\cos t)(C\cos t + D\sin t).
\end{align*}
This is not a "limaçon" in the stated sense unless the linear transformation is a Euclidean motion (rotation about the origin or reflection about a line through the origin) followed by scaling about the origin: We must have
\begin{align*}
  A\cos t + B\sin t &= E\cos(t + t'), \\
  C\cos t + D\sin t &= E\sin(t + t')
\end{align*}
for some real $E$ and $t'$. The remaining details are a little tedious to write out, but follow from addition formulas for the trig functions, which lead to $A^{2} + B^{2} = E^{2} = C^{2} + D^{2}$ and $AC + BD = 0$, so that the transformation matrix has orthogonal columns of equal length.
As for the equations of the transformed curve, multiplying the equation of the limaçon by $r$ gives
$$
x^{2} + y^{2} = r^{2} = ar + br\cos t = ar + bx.
$$
Rearranging and squaring,
$$
(x^{2} + y^{2} - bx)^{2} = (ar)^{2} = a^{2}(x^{2} + y^{2}).
$$
This quartic equation in $(x, y)$ may be expressed in terms of $u$ and $v$ by noting that
\begin{align*}
  \left[\begin{array}{@{}c@{}}
      x \\
      y \\
    \end{array}\right]
  &= \left[\begin{array}{@{}cc@{}}
      A & B \\
      C & D \\
    \end{array}\right]^{-1}
  \left[\begin{array}{@{}c@{}}
      u \\
      v \\
    \end{array}\right] \\
  &= \frac{1}{AD - BC} \left[\begin{array}{@{}rr@{}}
      D & -B \\
      -C & A \\
    \end{array}\right]
  \left[\begin{array}{@{}c@{}}
      u \\
      v \\
    \end{array}\right] \\
  &= \frac{1}{AD - BC} \left[\begin{array}{@{}c@{}}
      Du - Bv \\
      Av - Cu \\
    \end{array}\right].
\end{align*}
The details are again a bit tedious, but the result is a quartic polynomial, and the zero locus has the same qualitative shape (e.g., a cusp or crossing at the origin becomes a cusp or crossing at the origin) even if it is not strictly a limaçon.
