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
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $7$. To find the pole of the line $\ldots$ with respect to the circle $\ldots$