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Reading a plane geometry book I found the following exercise:

Given a circle $K=(O,k)$ and a point $P$, the power of $P$ with respect to $K$ is the quantity $|OP|^2-k^2$. Let $P$ and $Q$ be conjugates with respect to $K$. Show that $|PQ|^2$ is the sum of the powers of $P$ and $Q$ with respect to $K$.

Note: Given a circle $K$ and two points $P$ and $Q$, we say they are conjugate points with respect to $K$ if the polar of each point passes through the other point.

I was thinking about using the pythagorean theorem and the definition of inverse point. However, I didn't get the desired result.

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Let $P'$ be the inverse of $P$ w.r.t. $K$; it is furthermore well-known that $P'$ lies on the polar of $P$, just as $Q$ does. Therefore $\angle QP'O=\angle PP'Q=90^\circ$, so we can use the Pythagorean theorem to conclude that $$PQ^2-P'P^2=P'Q^2=QO^2-P'O^2$$\begin{align*}\implies PQ^2- \lvert PO-P'O\rvert^2 &=QO^2-P'O^2\\\implies PQ^2-\left(PO-\frac{r^2}{PO}\right)^2&=QO^2-\frac{r^4}{PO^2}\\\implies PQ^2&=QO^2+PO^2-2r^2\\&=(QO^2-r^2)+(PO^2-r^2)\end{align*}

enter image description here

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Edit: Corrected proof.

We use cartesian coordinates in the plane, so that $O$ is the origin. Let $P$ be on the $Ox$ axis, $P=(x,0)$ for some $x\ne 0$.

The polar is the vertical line passing through $P^*=(k^2/x,0)$, so $Q$ is a point of the shape $Q=(k^2/x,y)$ for some real $y$.

We have to check: $$ \left(x-\frac {k^2}x\right)^2+y^2 =(x^2-k^2)+\left(\frac {k^4}{x^2}+y^2-k^2\right)\ . $$ Yes.

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  • $\begingroup$ $O,P,Q$ don't have to be collinear... $\endgroup$
    – Dr. Mathva
    Oct 11 '20 at 23:14
  • $\begingroup$ What are conjugates?! @Dr.Mathva Conjugates w.r.t. inversion? $\endgroup$
    – dan_fulea
    Oct 11 '20 at 23:18
  • $\begingroup$ They just need to lie on the others polar ... the conjugate after inversion is only a very special case. Have a look here $\endgroup$
    – Dr. Mathva
    Oct 11 '20 at 23:20
  • $\begingroup$ ok, the definition uses the polars... sorry... $\endgroup$
    – dan_fulea
    Oct 11 '20 at 23:20

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