There is a problem on the book Introduction to Partial Differential Equations with Applications by E. C. Zachmanoglou and Dale W. Thoe on page 240:
Show that inversion with respect to $ S(0,1) $ maps the quarter plane $ \Omega=\{(x,y)\in \mathbb{R}^2,x>1,y>1\} $ onto the lens-shaped domain bounded by the two circles $$\left(x-\dfrac{1}{2} \right)^2+y^2=\dfrac{1}{4},~~ x^2+ \left(y-\dfrac{1}{2} \right)^2=\dfrac{1}{4} .$$ But I don't know how to show. Any help?
The region in question is bounded by the lines $\ell_1=\{(t,1):t\in\mathbb{R}\}$ and $\ell_2=\{(1,t):t\in\mathbb{R}\}$.
Inversion takes a point $p=(t,1)$ on $\ell_1$ to $\frac{1}{|(t,1)|^2}(t,1)=\left(\frac{t}{t^2+1},\frac{1}{t^2+1}\right)$
The $x,y$ coordinates of this image point $p'$ satisfy the equation $\frac{x^2}{y^2}=t^2=\frac{1}{y}-1$ and so the image of $\ell_1$ is $\ell_1'=\{(x,y):x^2+y^2=y\}$. Similarly, by interchanging the role of $x,y$ we get that the image of $\ell_2$ is $\ell_2'=\{(x,y):x^2+y^2=x\}$.
These are the two circles in your question. Moreover, inversion swaps $0$ and $\infty$, so the image of $\{x>1\}$ and $\{y>1\}$ must be the $\textit{interior}$ of the two circles $\ell_1',\ell_2'$ respectively. The resulting intersection of images is thus the lens shape you describe.