# Harmonic Property of Pole and Polar/Circle

## Theorem

Let $\CC$ be a circle whose radius is $r$ and whose center is at the origin of a Cartesian plane.

Let $P$ be an arbitrary point in the Cartesian plane.

Let $\LL$ be a straight line through $P$ which intersects $\CC$ at points $U$ and $V$.

Let $Q$ be the point where $\LL$ intersects the polar of $P$.

Then $\tuple {PQ, UV}$ is a harmonic range.

## Proof

From Equation of Circle center Origin, we have that the equation of $\CC$ is:

$x^2 + y^2 = r^2$

Let $P = \tuple {x_1, y_1}$.

Let $Q = \tuple {x, y}$ be a point on $\LL$.

Let $V$ divide $PQ$ in the ratio $k : 1$.

Then the coordinates of $V$ are:

$V = \tuple {\dfrac {k x + x_1} {k + 1}, \dfrac {k y + y_1} {k + 1} }$

Substituting into the equation for $\CC$:

 $\ds \paren {k x + x_1}^2 + \paren {k x + y_1}^2$ $=$ $\ds r^2 \paren {k + 1}^2$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds k^2 \paren {x^2 + y^2 - r^2} + 2 k \paren {x x_1 + y y_1 - r^2} + \paren { {x_1}^2 + {y_1}^2 - r^2}$ $=$ $\ds 0$

The roots of this quadratic equation in $k$ correspond to the points $U$ and $V$.

Now let $Q$ lie on the polar of $P$. Then by definition of polar:

$x x_1 + y y_1 - r^2 = 0$

and the roots of $(1)$ are equal but of opposite sign.

That is, $U$ and $V$ divide $PQ$ internally and externally in the same ratio.

Hence $\tuple {PQ, UV}$ is a harmonic range.

$\blacksquare$