I am given $y_0\neq 0$ and $$\left\{u+iv=\frac{1}{x+iy_0}:x\epsilon R\right\}$$
I'm asked to find a geometric description of this. I notice that $w=\frac{1}{z}$ so $z=\frac{1}{w}$ and if $z=x+iy$ and $w=u+iv$ then $u=\frac{x}{x^2+y^2}$, $v=\frac{-y}{x^2+y^2}$ and on the reverse we have $x=\frac{u}{u^2+v^2}$ and $y=\frac{-v}{u^2+v^2}$
From here I'm pretty much stuck. I know $y=y_0$ so $\frac{-v}{u^2+v^2}=y_0$ but I have no idea if this helps me
Given $y_0 \in \Bbb{R}$, let $z=x+iy_0$, then we have $$v=\frac{-y_0}{|z|^2}$$ We also have $|w|\,|z|=1$. Thus $\frac{1}{|z|^2}=u^2+v^2$. Using this in the second equation given above, we get (which is also what you have stated as one of your derivations) $$v=-y_0(u^2+v^2) \implies y_0u^2+y_0v^2+v=0$$
Since $y_0 \neq 0$, then we can write this as $$u^2+\left(v+\frac{1}{2y_0}\right)^2=\frac{1}{4y_0^2}.$$ This is a circle with center at $\left(0,\frac{-1}{2y_0}\right)$ and radius $\frac{1}{2y_0}$.