Complex Analysis (Q on Riemann Sphere)

Let $\displaystyle f(z) = \frac{1-i}{z-i}, z \neq i$ and $\displaystyle \phi: \mathbb{C} \to \mathbb{S}$ be the stereographic projection of $\displaystyle \mathbb{C}$ into the Riemann Sphere $\displaystyle \mathbb{S}$.

Explain why the image $\displaystyle \phi \circ f(C)$ in $\displaystyle \mathbb{S}$ of the set

$\displaystyle C = \{ z \in \mathbb{C}: z \neq i, |z-1-i | =1\}$

by the map $\displaystyle \phi \circ f$ lies on a circle through $\displaystyle \infty $, the north pole of $\displaystyle \mathbb{S}$.

Is it possible to do this question without finding f(C) first?