Geometrical proof for property of chord of contact with a circle Let their be a circle with center O, and radius r. Let their be a point P outside the circle at a distance d from O. Let a variable line passing through P cut the circle in 2 points, A and B. We define a point C between A and B such that PC is harmonic mean of PA and PB. Prove that the locus of C is chord of contact of P with respect to the circle.
I can do it through coordinate geometry, I just wanted to find a geometrical method. I tried looking at its inversion, and trying to find relations but I couldn't.
Any help is appreciated. Thanks.
 A: 
Let the circle have centre $O$.Let the points at which the tangents touch the circle be $D$ and $E$.A variable line through point $P$ cuts the circle at points $A$ and $B$. $DE$ cuts this line at $C$. So proving that $PC$ is the harmonic mean of $PA$ and $PB$ will prove the given statement.
Let $M$ be the midpoint of $AB$. Then $OM\perp AP$.
Now, observe that, pentagon $MODPE$ is cyclic. Hence, $PC\cdot CM=DC\cdot CE$.
Also, since $DC\cdot CE=AC\cdot CB$, $AC\cdot CB=PC\cdot CM$.
$AC=PA-PC$
$BC=PC-PB$
$CM=\frac {1}{2}(AC-BC)=\frac {1}{2}(PA+PB)-PC$
Plugging these values into the equation gives,
$PC\{\frac {1}{2}(PA+PB)-PC\}=(PA-PC)(PC-PB)$
$\Rightarrow   PC=\frac {2PA\cdot PB}{PA+PB}$
Hence,  $PC$ is the harmonic mean of $PA$ and $PB$.
A: 
Another approach. For simplicity let p has distance 2R from the center of circle with center at O and radius R. If PC is tangent at C We have A, B and C coincident and we may write:
$PC=PA=PB$
For any position of A and B we have:
$PC^2=PA\times PB$
$PC=\frac {PA\times PB}{PC}=\frac {2\times PB\times PA }{2 PC=PA+PB}$
Now we evaluate the position of point H which is the foot of altitude of right angled triangle PCO.
$CP^2=OP^2-CO^2=(2R)^2-R^2=3R^2$
We have area of this triangle as:
$S=\frac{PC \times CO}2=\frac{R^2\sqrt 3}2$
$CH=\frac{R^2\sqrt 3}{2\times 2R}=\frac{R\sqrt 3}4$
In triangle PCO we can write:
$CH^2=PH\times OH$
$PH(OP-PH)=\frac{3R^2}4$
OR:
$2R\times PH-PH^2=\frac{3R^2}4$
$4PH^2+3R^2-8PH\times R=0$
Solving this equation for PH we get:
$PH=\frac {3R}2=\frac{6R^2}{4R}=\frac{2(R)\cdot (3R)}{R+3R}$
Or:
$PH=\frac{2 PA\times PB}{PA+PB}$
Which means H has similar particularity as C and CH is part of chord CC'.
