Condition for Lines to be Conjugate
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Theorem
Let $\CC$ be a circle of radius $r$ whose center is at the origin of a Cartesian plane.
Let $\PP$ and $\QQ$ be conjugate lines with respect to $\CC$:
| \(\ds \PP: \ \ \) | \(\ds l_1 x + m_1 y + n_1\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \QQ: \ \ \) | \(\ds l_2 x + m_2 y + n_2\) | \(=\) | \(\ds 0\) |
Then:
- $l_1 l_2 + m_1 m_2 = \dfrac {n_1 n_2} {r^2}$
Proof
By definition of conjugate lines, $\PP$ and $\QQ$ are the polars of points $P$ and $Q$ respectively, such that $P$ lies on $\QQ$ and $Q$ lies on $\PP$.
From Coordinates of Pole of Given Polar, $P$ is given by:
- $P = \tuple {-\dfrac {l_1} {n_1} r^2, -\dfrac {m_1} {n_1} r^2}$
We have that $P$ lies on $\QQ$.
Substituting $x = -\dfrac {l_1} {n_1} r^2$ and $y = -\dfrac {m_1} {n_1} r^2$ in the equation of $\QQ$, we obtain:
| \(\ds l_2 \paren {-\dfrac {l_1} {n_1} r^2} + m_2 \paren {-\dfrac {m_1} {n_1} r^2} + n_2\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds l_1 l_2 r^2 + m_1 m_2 r^2\) | \(=\) | \(\ds n_1 n_2\) | multiplying by $n_1$ and rearranging |
from which the result follows.
$\blacksquare$
Sources
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $8$. Reciprocal property of pole and polar