# strong deformation retract question

I'm just learning about homotopy equivalent spaces, and I'm having a little trouble with an exercise: Let $x_0\in\mathbb{R^2}$. Find a circle in $\mathbb{R^2}$ which is a strong deformation retraction of $\mathbb{R^2}-${$x_0$}.

If I understand the def. of strong deformation retract correctly, intuitively, any circle containing $x_0$ will do: You push the space inside the circle to the circle and pull the space outside the circle to the circle.

My attempt: Send $x_0$ to $0$ in the coordinates $(x',y')$. Then take a circle $C'$ of raidus $k$ centered at $0$ in these primed coordinates. For a fixed $\theta'$ send $(r',\theta')$to $(k,\theta')$. This sends everything in $\mathbb{R^2}-${$0$} to $C'$ and fixes every point $c'\in C'$, so $C'$ is a strong deformation retract of $\mathbb{R^2}-${$0$}. Then rewrite in the original $(x,y)$ coordinates.

Your answer is correct. Easier to describe is to say that the circle is the set satisfying the equation $(x-h)^2+(y-k)^2=1$ where $x_0=(h,k)$. And the strong deformation retract is: $$H_t(x)=\left(1-t+\frac{t}{\lVert x-x_0 \rVert}\right)\cdot(x-x_0)$$ Geometrically, as $t \to 1$ it takes a point $x$ towards the intersection between the unique line passing through $x$ and $x_0$ and the unitary circle. If $t=0$ it is just a linear translation of the plane in the direction of $x$. Thus $H$ combines a translation always proportional to $x$ and an increasing contraction/expansion towards the circle.