Coordinates of Pole of Given Polar

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Theorem

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

Let $\LL$ be a straight line whose equation is given as:

$l x + m y + n = 0$

Then the pole $P$ of $\LL$ with respect to $\CC$ is:

$P = \tuple {-\dfrac l n r^2, -\dfrac m n r^2}$


Homogeneous Coordinates

The pole $P$ of $\LL$ with respect to $\CC$, given in homogeneous Cartesian coordinates is:

$P = \tuple {l, m, -\dfrac n {r^2} }$


Proof

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

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

Let $P = \tuple {x_0, y_0}$.

By definition of polar:

$x x_0 + y y_0 = r^2$

Comparing this with the equation for $\LL$:

$\dfrac {x_0} l = \dfrac {y_0} m = \dfrac {r^2} {-n}$

The result follows.

$\blacksquare$


Sources