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.

  • $\begingroup$ What do you mean by ' chord of contact of P with respect to the corcle'? $\endgroup$
    – sirous
    Jan 24, 2021 at 9:25
  • 1
    $\begingroup$ @sirous You can draw two tangents from $P$ to the given circle. If these two tangents touch the circle at $D$ and $E$, then the chord of contact of $P$ wrt the said circle is $DE$. $\endgroup$
    – YNK
    Jan 24, 2021 at 9:48

2 Answers 2


Circle with centre O

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 $DC\cdot CE=AC\cdot CB$, $AC\cdot CB=PC\cdot CM$.



$CM=\frac {1}{2}(AC-BC)=\frac {1}{2}(PA+PB)-PC$

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 $PB$.

  • 1
    $\begingroup$ $OMDPE$ is not a pentagon, but $MODPE$ is. $\endgroup$
    – YNK
    Jan 24, 2021 at 15:26
  • $\begingroup$ @YNK Thanks for pointing it out. $\endgroup$
    – Limestone
    Jan 24, 2021 at 15:42

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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:


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.


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:

$CH^2=PH\times OH$



$2R\times PH-PH^2=\frac{3R^2}4$

$4PH^2+3R^2-8PH\times R=0$

Solving this equation for PH we get:

$PH=\frac {3R}2=\frac{6R^2}{4R}=\frac{2(R)\cdot (3R)}{R+3R}$


$PH=\frac{2 PA\times PB}{PA+PB}$

Which means H has similar particularity as C and CH is part of chord CC'.


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