Let $L$ be a circle in the Riemann sphere such that $L$ and the circle $\partial \Delta(1)$ are perpendicular. If $f \in Aut(\Delta(1))$ is a linear transformation then are $f(L)$ and $\partial\Delta(1)$ are orthogonal?
Attempt:
I think it it's true. First let $L$ be the circle $|z-b|=r$ for $b\in \mathbb{C} $. Since $L$ and $\partial\Delta(1)=\{z\in\mathbb{C}:|z|=1\}$ are orthogonal we have $$r^2+1=|b|^2$$
Since $f \in Aut(\Delta(1))$ then $f$ can be written as $f=e^{i\theta}\frac{z-a}{1-\overline{a}z}$ where $a\in\Delta(1)$ and $\theta\in \mathbb{R}$. I am just confused what the image of $L$ would be under $F$. After finding the image I would like to use the relation between $r$ and $b$ to prove $f(L)$ and $\partial\Delta(1)$ are orthogonal. Am I doing this right or is there another approach ? Also would like to get some help on finding the image of $L$ under $f$.
This transformation is conformal and the image of the unit circle is the unit circle. That shows that the images of two orthogonal circles is orthogonal.