# Proving $|PQ|^2$ is the sum of the powers of $P$ and $Q$ with respect to a circle

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.

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*}

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.

$$\square$$

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