Consider the family $$C_{\alpha}:\big({2(\alpha-\cos{t})\,\cos{t},\,2(\alpha-\cos{t})\,\sin{t}}\big)\,,\;t\in[0,2\pi]\,,\;\alpha\in\mathbb{R},$$ of closed parametric curves. Let $X_{\alpha}$ is the image set of $C_{\alpha}$ equiped with the induced topology of $\mathbb{R}^2$.
- For which values of $\alpha$ these topological spaces $X_{\alpha}$ are homotopy-equivalent to topological space $X_{0.5}$?
- For those spaces which are homotopy-equivalent to $X_{0.5}$ provide a homotopy equivalence.
In the figure below depicted $6$ curves of the family $C_{\alpha}$ for $\alpha\in\{{0.5,0.6,1.0,0.3,1.3}\}$.
Note: This is not a homework exercise. I've got an outline of solution.
Edit: I had to clarify my question. Thanks to John Palmieri's remarks.
I give an outline of solution:
To handle the problem easier, for $\alpha\in({-1,1})$ we consider the below differentiable parametrization of the given curves: \begin{align*} R_{\alpha}&:[0,2\pi]\longrightarrow\mathbb{R}^2\,;\\ &t\longmapsto R_{\alpha}(t)=\begin{pmatrix} 2\big({\alpha-\cos({t+\arccos{\alpha}})}\big)\,\cos({t+\arccos{\alpha}})\\ 2\big({\alpha-\cos({t+\arccos{\alpha}})}\big)\,\sin({t+\arccos{\alpha}}) \end{pmatrix}\,. \end{align*} We observe that $R_{\alpha}(0)=R_{\alpha}(2\pi)=(0,0)$. Also, for every $\alpha\in({-1,1})$, the curve $R_{\alpha}$ makes two rounds with intersection point at $(0,0)$. For $\alpha\in({-\infty,-1}]\cup [{1,+\infty})$ the curve $C_{\alpha}$ is simple closed continuous curve.
We claim that the space $X_{\alpha}$ is homotopy-equivalent to space $X_{0.5}$, iff $\alpha\in({-1,1})$.
First we claim that, for $\alpha\in({-\infty,-1}]\cup [{1,+\infty})$ doesn't exist a continuous function $f_{\alpha}:X_{0.5}\longrightarrow X_{\alpha}$, because if exists then must exist an open connected set $A$ of $X_{\alpha}$ (i.e. a piece of curve) such that $f^{-1}_{\alpha}(A)$ is open connected set $Α$ of $X_{0.5}$ with $(0,0)\in f^{-1}_{\alpha}(A)$, which cannot happened because $(0,0)$ is an intersection point of $X_{0.5}$ ($f^{-1}_{\alpha}(A)$ is the union of two crossed pieces of curves).
Secondly, for $\alpha\in({-1,1})$, we observe that the maps $f_{\alpha}:X_{0.5}\longrightarrow X_{\alpha}$ with \begin{align*} &\begin{pmatrix} 2\big({0.5-\cos({t+\arccos{0.5}})}\big)\,\cos({t+\arccos{0.5}})\\ 2\big({0.5-\cos({t+\arccos{0.5}})}\big)\,\sin({t+\arccos{0.5}}) \end{pmatrix} \longmapsto\\ &\hspace{1.5cm} f_{\alpha}\big({x(t),y(t)}\big)=\begin{pmatrix} 2\big({\alpha-\cos({t+\arccos{\alpha}})}\big)\,\cos({t+\arccos{\alpha}})\\ 2\big({\alpha-\cos({t+\arccos{\alpha}})}\big)\,\sin({t+\arccos{\alpha}}) \end{pmatrix}\,,\;t\in[0,2\pi)\,, \end{align*} and
$g_{\alpha}:X_{\alpha}\longrightarrow X_{0.5}$ with \begin{align*} &\begin{pmatrix} 2\big({\alpha-\cos({t+\arccos{\alpha}})}\big)\,\cos({t+\arccos{\alpha}})\\ 2\big({\alpha-\cos({t+\arccos{\alpha}})}\big)\,\sin({t+\arccos{\alpha}}) \end{pmatrix}\longmapsto\\ &\hspace{1.5cm} g_{\alpha}\big({u(t),v(t)}\big)=\begin{pmatrix} 2\big({0.5-\cos({t+\arccos{0.5}})}\big)\,\cos({t+\arccos{0.5}})\\ 2\big({0.5-\cos({t+\arccos{0.5}})}\big)\,\sin({t+\arccos{0.5}}) \end{pmatrix}\,,\;t\in[0,2\pi)\,, \end{align*} are continuous* -it's sufficient to observe that $f_{\alpha}(0,0)=(0,0)$ and $g_{\alpha}(0,0)=(0,0)$- with $f_{\alpha}\circ g_{\alpha}={\rm{id}}_{X_{\alpha}}$ and $g_{\alpha}\circ f_{\alpha}={\rm{id}}_{X_{0.5}}$. Therefore, for every $\alpha\in({-1,1})$, the pair $({f_{\alpha}, g_{\alpha}})$ is a homotopy-equivalence.
(*) In fact they are homeomorphisms with $f^{-1}_{\alpha}= g_{\alpha}$.