General proof of the Power of a Point Theorem (uncomplicated) The definition of The Power of a Point Theorem in the Wolfram Demonstrations Project is the following:

If two chords $AB$ and $CD$ in a circle intersect the point $P$, then $|AP||PB|=|CP||PD|$

So far, all the proofs of this theorem I have found are restricted to cases: $P$ is either necessarily inside or outside the circle. What would be an uncomplicated proof of this theorem comprising both cases at once?
 A: In the most-nuanced interpretation, the "power" of a point $P$ with respect to a circle $\gamma$ is a signed value:
$$\operatorname{pow}_\gamma (P) = p^2-r^2 \tag{1}$$
where $r$ is the circle's radius, and $p$ is the distance from $P$ to the circle's center. This power value is positive for $P$ outside $\gamma$ (ie, $p>r$), negative for $P$ inside ($p<r$), and zero for $P$ on the circumference ($p=r$). This trichotomy typically invites breaking the proof that "the power equals product of distances from $P$ to collinear intersections with $\gamma$" into cases, to avoid introducing "signed distances" into the argument.

Coordinate geometry doesn't discriminate against signed distances, so let's take a look at a quick argument there. Let $\gamma$ be centered at the origin, so that its equation is 
$$x^2+y^2=r^2 \tag{2}$$ 
Let $P$ have coordinates $(p,0)$, and suppose a line through $P$ makes an angle $\theta$ with the $x$; we can describe the points on this line by
$$P + t\,(\,\cos\theta, \sin\theta\,) \tag{3}$$
where $t$ amounts to the signed distance along that line $P$. (Points in one direction along the line have positive distances; points in the other direction have negative distances.) To find the points where the line meets the circle, we merely substitute the coordinates from $(3)$ into $(2)$.
$$(p+t\cos\theta)^2 + (0+t\sin\theta)^2 = r^2 \qquad\to\qquad t^2 + 2 p t \cos\theta + p^2-r^2 = 0 \tag{4}$$
By Vieta's formula (or by solving the quadratic and multiplying), we find that the product of the roots of $(4)$ is the constant term. This says exactly that 

The product of the signed distances from $P$ to collinear points of intersection with $\gamma$ is $p^2-r^2$ (which is to say, the power of $P$ with respect to $\gamma$).

The reader should note that, if $P$ is inside the circle, then $P$ is between the points of intersection; thus, those points have oppositely-signed distances from $P$, making the product of those distances negative. Contrariwise, if $P$ is outside the circle, the points have same-signed distances, making the product positive. Finally, for $P$ on the circle, at least one of those distances is $0$, making the product $0$, too.
A: The power of a circle $ x^2+y^2+2fx+2gy+P=0$  is $P$ with respect to the origin.
It is positive if the origin/pole is outside and negative if inside ( according as $ (h>a,\, h<a\,) $  and zero if on the boundary. It is obtained when we set $x=0,y=0 $ intact with its sign in the above circle equation.
The product of line segments ( Euclidean definition) in a circle is signed as per a convention.
The definition you have given is for absolute values of length segments and stated so as to include both signs.
The definition of sign of power $P$ follows a convention, taken negative when pole/origin is inside the boundary. No need of a proof for any convention.It can be extended to other curves also.
We can draw analogies from physics. In Electrostatics for example we have a set of orthogonal $(\tau,\sigma) $ (red,blue) Bipolar Coords as Apollonian Circles representing lines of force and equipotential lines originating from real and imaginary parts of a single function of a complex variable $ w= \coth (z)$ placed with respect to a common origin and an arbitrary sign for their power.
In Mechanics too Mohr curvature Circle representation of Gauss curvature is of positive sign when both principal curvatures translate as positive power and negative total curvature product for principal curvatures of opposite sign. (Synclastic/Anticlastic geometries).
Hope this does not complicate the matter further :) ..

