# 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.

• What do you mean by ' chord of contact of P with respect to the corcle'? Jan 24, 2021 at 9:25
• @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 $DE$.
– YNK
Jan 24, 2021 at 9:48

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 $$PB$$.

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

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'.