What's the u-v equation for u and v defined as follows? I can not solve the u-v equation for $u=\frac{1-r^2}{1-2r \cos \theta + r^2}$ for $0 < r < 1$ and $v=\frac{2r\sin\theta}{1-2r \cos \theta + r^2}$ for $0 < r < 1.$  I assume that it is a circle because it is the polar form of a bilinear function $ w=\frac{i(1-z)}{1+z}.$ I have tried to equate $u^2+(v-t)^2=k$, for some t and k. And figure out what is $t$ and $k$. But I failed. Is there any suggestion ? Thank you!

Your expressions should read
\begin{align}
  w &= \frac{i(1-z)}{1+z} \\
  u+vi &= \frac{2r\sin \theta+i(1-r^2)}{1+2r\cos \theta+r^2} \\
  \color{red}{u} &= \frac{2r\sin \theta}{1\color{red}{+}2r\cos \theta+r^2} \\
  \color{red}{v} &= \frac{1-r^2}{1\color{red}{+}2r\cos \theta+r^2}
\end{align}

  
*
  
*$\theta=0 \implies
 (u,v)=\left( 0, \dfrac{1-r}{1+r} \right)$
  
*$\theta=\pi \implies
 (u,v)=\left( 0, \dfrac{1+r}{1-r} \right)$
  
*$2t=\dfrac{1-r}{1+r}+\dfrac{1+r}{1-r} \implies t=\dfrac{1+r^2}{1-r^2}$
  
*$2\sqrt{k}=\dfrac{1+r}{1-r}-\dfrac{1-r}{1+r} \implies \sqrt{k}=\dfrac{2r}{1-r^2}$
  
*Try to verify $$u^2+\left(v-\frac{1+r^2}{1-r^2} \right)^2=\frac{4r^2}{(1-r^2)^2}$$

