Understanding the formula for stereographic projection of a point. I was wondering about the equation of line I can write which can help me finding the coordinates of Point $P'$ in relation with coordinates of points on the sphere that is $P$.

Let the $P'(X,Y)$, then how can I find these following coordinates -
I tried equating the slope of $NP'$ with that of the slope of $PP'$.
but could only write $\frac{y}{x} = \frac{Y-y}{X-x}$.
I donot know how it is getting these coordinates -
$(X,Y) = (\frac{2ax}{a-z},\frac{2ay}{a-z})$
and $(x,y,z) = (\frac{2aX}{X^2+Y^2+a^2},\frac{2aY}{X^2+Y^2+a^2},\frac{X^2+Y^2-a^2}{X^2+Y^2+a^2})$?
Here we are following this notation that - points lying on the sphere are represented by $(x,yz)$ and points lying outside the sphere by $X,Y$
 A: Let $S$ be the point of tangency of the sphere and the plane,
and let $Q$ lie on line $NS$ such that angle $\angle NQP$ is a right angle.

Observe that triangles $\triangle NQP$ and $\triangle NSP'$ are similar,
with $NQ = a - z$ and $NS = 2a.$
Therefore $$\frac{P'S}{PQ} = \frac{2a}{a-z}. \tag1$$
But we also have $P = (x,y,z)$ while $Q = (0,0,z),$
and $P' = (X,Y,-a)$ (in three dimensions) while $S = (0,0,-a).$
By proportions,
$$\frac Xx = \frac Yy = \frac{P'S}{PQ},$$
and therefore (using equation $(1)$ to substitute for $\frac{P'S}{PQ}$),
$$
X = \frac{2a}{a-z}x \qquad\text{and}\qquad Y = \frac{2a}{a-z} y.
$$
To transform coordinates in the other direction, observe that
triangles $\triangle NPS$ and $\triangle NSP'$ are similar, with
$$
\frac{PN}{NS} = \frac{NS}{P'N}.
$$
Then
$$
\frac xX = \frac{PQ}{P'S} = \frac{PN}{P'N} = \frac{(NS)^2}{(P'N)^2},
$$
so
$$
x = \frac{(NS)^2}{(P'N)^2}X.
$$
Seeing that $P'N$ is the hypotenuse of a right triangle with legs
$\sqrt{X^2 + Y^2}$ and $2a,$
so $$(P'N)^2 = X^2 + Y^2 + 4a^2,$$
and recalling that $NS = 2a,$ we have
$$
x = \frac{4a^2}{X^2 + Y^2 + 4a^2}X. \tag2
$$
Similarly,
$$
y = \frac{4a^2}{X^2 + Y^2 + 4a^2}Y. \tag3
$$
Using the fact that $z^2 = a^2 - x^2 - y^2,$
using equations $(2)$ and $(3)$ to substitute for $x$ and $y,$
and performing some algebraic manipulations, we find that
$$
z = \frac{a(X^2 + Y^2 - 4a^2)}{X^2 + Y^2 + 4a^2}.
$$
The formulas you were trying to justify for $(x,y,z)$ are then seen
not to be correct.
Those formulas, multiplied by $a,$ would give $(x,y,z)$ in terms of the
coordinates of $P$ projected onto the plane $z = 0$
(not $z = -a$); that may explain where those formulas came from
(though this does not explain the missing factor of $a$).
If you multiply each of the formulas
$\frac{2aX}{X^2+Y^2+a^2},$ $\frac{2aY}{X^2+Y^2+a^2},$
and $\frac{X^2+Y^2-a^2}{X^2+Y^2+a^2}$ by $a,$ 
substitute $\frac X2$ for $X,$ and substitute $\frac Y2$ for $Y,$
you will have correct formulas for the reverse projection from
the plane $z = -a.$
A: Any point on the line joining N and P is of the type $(x,y,z)+\lambda (x,y,z-a)$ [ Here $(x,y,z-a)=(x,y,z)-(0,0,a)$]. We have to choose $\lambda $ such that the last coordinate $z+\lambda (z-a)$ becomes $-a$. This gives $\lambda =\frac {a+z} {a-z}$. Now simply calculate the first two coordinates using this value of $\lambda$. For example the first coordinate is $x+\lambda x =(1+\frac {a+z} {a-z})x=\frac {2a} {a-z} {x}$. Similarly, the second coordinate is $y+\lambda y =(1+\frac {a+z} {a-z})y=\frac {2a} {a-z} {y}$.
