Find a Möbius transformation $T$ from $\{z=x+iy:x+y>0\}$ onto the disk $D(1,4)$, such that $T(1)=2$ and $T(0)=-3$.
The proof given is as follows:
Since $-i$ is symmetric to $1$ with respect to the line $x+y=0$, then $T(1)$ and $T(-i)$ is symmetric with respect to $C(1,4)$, the circle centred at $1$ with radius $4$. This gives $T(-i)=17$. Since $T$ preserves cross-ratio, $(T(z),-3,2,17)=(T(z),T(0),T(1),T(-i))=(z,0,1,-i)$. Simplifying, we get \begin{equation*} T(z)=\frac{(17-8i)z-9}{(1-4i)z+3}. \end{equation*}
My question:
I know that every Möbius transformation preserves symmetry in the sense that if $z$ is symmetric to $z^{\ast}$ with respect to a circle or line $C$, then $T(z)$ and $T(z^{\ast})$ is symmetric with respect to $T(C)$.
In the above proof, how do we know that $T(C)=T(L)$ is precisely $C(1,4)$, where $L$ is the line $x+y=0$?
More generally, how can we determine an image of circle or line under a Möbius transformation which we are supposed to find, suppose that we know some points of symmetry (as in the above)?