A circle with center C is cut by a line through origin O at P and Q. If M is mid-point of PQ, show that $ OM^2 - MP^2 $ is constant for all inclinations of $OP$ and equals its power ( square of tangent ) from O.
EDIT1:
The motivation in posting is that in my present view this geometrical result should rank at par with the Pythagorean theorem:
$$ a^2 + b^2 = const., \,\, c^2 - d^2 = const. \, $$
and that it may have significance in hyperbolic geometry.
On
WLOG the equation of the circle $$(x-a)^2+(y-b)^2=r^2$$ and the equation of the straight line through origin $$y=mx$$
If $P(x_1,y_1),Q(x_2,y_2); M\left(\dfrac{x_1+x_2}2,\dfrac{y_1+y_2}2\right)$
$y_1=mx_1,y_2=mx_2$
$$\implies(x-a)^2+(mx-b)^2=r^2\iff(1+m^2)x^2-2x(a+mb)+a^2+b^2-r^2=0$$
$x_1\cdot x_2=\dfrac{a^2+b^2-r^2}{1+m^2}$
$$OM^2=\left(0-\dfrac{x_1+x_2}2\right)^2+\left(0-\dfrac{y_1+y_2}2\right)^2=\dfrac{(1+m^2)(x_1+x_2)^2}4$$
$$MP^2=\left(x_1-\dfrac{x_1+x_2}2\right)^2+\left(y_1-\dfrac{y_1+y_2}2\right)^2=\dfrac{(1+m^2)(x_1-x_2)^2}4$$
Can you take it from here?
This is definitional if you factor $OM^2 - MP^2$ as a difference of two squares. Power can be defined as $OP \times OQ$ or as square-of-tangent, different books do it differently, but most books prove that the two quantities are the same.