A special Mobius Transformation that maps the right half plane to the unit disc

Find the Mobius Transformation that maps the right half plane to the unit disc carrying the point $z=15$ to the origin.

Since the Mobius transformation takes the point $z=15$ to the origin, so I can say it has the following form

$$T(z)=k\frac{z-15}{z-c}, \:\:\:\text{where k,c are complex numbers}.$$

Now I know that Mobius transformation preserves the symmetry. I need to find some symmetric points with respect to upper half plane.

If I am not mistaken I think any pure imaginary numbers $iy$ and $-iy$ are symmetric with respect to upper half plane ( I'm not sure about this statement though). I appreciate if you help me to learn which two points are symmetric in right half plane.

By symmetry, you can first say that $T(0)$ should be $-1$, so $\frac{-15k}{-c}=-1$, ie $c = -15k$. In particular, $T(z) = \frac{z - 15}{z/k + 15}$. Similarly, $T(\infty)$ should map to $1$, so $k=1$ and $T(z) = \frac{z-15}{z+15}$. Note that neither $T(0) = -1$ nor $T(\infty)=1$ were necessary to produce this map, but a Mobius map is defined on three points, and those seem like sensible choices.
It is not hard to verify this map satisfies the requirements also: Simply check the image of the imaginary axis is the unit circle, and since $15$ maps to $0$, we know the right half-plane maps to the unit disc.
This map is not unique: any post-multiplication by $e^{i \theta}$ for $\theta \in \mathbb{R}$ gives another valid map, for instance.