Where should I use conformality here? I am stuck in the following problem:

Given the Möbius transform $T(z)=\frac{z-9i}{z-i}$. Using conformality, compute and draw the image of the circle $x^2 +(y-2)^2=1$ using the fact that this circle is perpendicular to the imaginary axis.

I think the point is to apply the Möbius transformation on both imaginary axis and circle. I know that:

*

*A line passing through the origin in the $z$-plane is transformed in a line passing  through the origin in the $w$-plane.


*A circle passing through the origin in the in the $z$-plane is transformed into a line that does pass through the origin in the $w$-plane.
But I don't see how I can use conformality here.
 A: A map being conformal means that it preserves angles between smooth curves. That is, if the given circle intersects the imaginary axis perpendicularly, then the circle's image will intersect the imaginary axis' image perpendicularly. Since the (extended) imaginary axis is mapped to the (extended) real axis, and the original circle intersects the imaginary axis at $\mathrm i$ and $3\mathrm i$, we're looking for the generalized circle (=circle or line) that perpendicularly intersects the extended real axis at $T(\mathrm i)$ and $T(3\mathrm i)$.
A: You seem to be somewhat confused.  Lines in the plane are the circles  on the Riemann sphere that pass through $\infty$.  I think you must be looking at an example where the denominator in the transformation is $z$.  In this  case, the denominator is $z-i$ so that $T(i)=\infty$.  Any line or circle passing through $i$ will be transformed to a line, and any line or circle that does not pass through $i$ will be transformed to a circle.
You are right in that you have to find the images of the imaginary axis and the circle.  Where conformality comes in is that the two curves are orthogonal, so their images are orthogonal.  When is a line orthogonal to a circle?
