How to compute a linear transformation which carries the circle $|z|=2$ into $|z+1|=1$? 
Find the linear transformation which carries the circle $|z|=2$ into $|z+1|=1$, the point $-2$ into the origin, and the origin into $i$. 

In order to find the linear transformation will I use the following formula?
$$(z_1−z_3)(z_2−z)(z_2−z_3)(z_1−z)=(w_1−w_3)(w_2−w)(w_2−w_3)(w_1−w)$$ 
Or am I wrong?
Similar example but not the same:

Find a linear transformation mapping the circle $|z|=1$ onto the circle $|w-5|=3$ and taking the point $z=i$ to $w=2$. 

Given, $|w-5|=3$ so $|(w-5)/3)|=1$. So now, we can use the transformation $z=(w-5)/3$ but $z=i$ is not mapped to $w=2$. So we use the transformation $iz=(w-5)/3$. So, $z=i \implies w=3i*i+5=2$.
How to start ?

Not sure if this is correct so please let me know
Let $C_1 = |z| = 2$, and $C_2 = |z + 1| = 1$. Since the origin is in $C_1$ but not in $C_2$, we need to make the circle go inside out. Thus we get $T = \frac{az + b}{cz + d}$.
We know $T = z*(\frac{z + b}{z + d})$ $z*$ be the reflection point of the origin. We let $R$ be the radius of $C_2$, and $\alpha$ be the center of $C_2$.
Now $\infty$ is the inflection point of $0$. Therefore $Z^* = \frac{R}{\bar{z} - \bar{\alpha}} + \alpha = \frac{1}{-i + 1} -1 = \frac{1 + i - 1}{-i + 1} = \frac{i}{1 - i} = \frac{-1}{1 - i} \times \frac{1 + i}{1 + i} = \frac{-1 +i}{1 + 1} = \frac{-1 + i}{2}$.
So $T = \frac{-1 + i}{2} \times \frac{z + b}{z + d}$. So since $T(-2) = 0$ we get $b = 2$. It remains to solve for $d$
. We get $T(0) = i = \frac{-2 + 2i }{2d}$ therefore $d = \frac{-1 + i}{i} = i + 1$. 
So $T = \frac{(-1 + i)z + 2}{2z + (i + 1)}$
