This illustrates the fact that $\mathbb{R}^2\setminus\text{point}$ is homotopy equivalent to $S^1$. The key ingredient is the map $h\colon [0,1]\times\mathbb{R}^2\setminus\text{point}\to\mathbb{R}^2\setminus\text{point}$ given by $$ h(t,(x,y)) = \left(1-t + \frac{t}{\sqrt{x^2+y^2}}\right) (x,y), $$ so that $h(0,(x,y))=(x,y)$ and $h(1,(x,y))=(x,y)/\sqrt{x^2+y^2}\in S^1$.