Condition for Circles to be Orthogonal
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Theorem
Let $\CC_1$ and $\CC_2$ be circles embedded in a Cartesian plane.
Let $\CC_1$ and $\CC_2$ be described by Equation of Circle in Cartesian Plane as:
\(\ds \CC_1: \, \) | \(\ds x^2 + y^2 + 2 \alpha_1 x + 2 \beta_1 y + c_1\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \CC_2: \, \) | \(\ds x^2 + y^2 + 2 \alpha_2 x + 2 \beta_2 y + c_2\) | \(=\) | \(\ds 0\) |
Then $\CC_1$ and $\CC_2$ are orthogonal if and only if:
- $2 \alpha_1 \alpha_2 + 2 \beta_1 \beta_2 = c_1 + c_2$
Proof
When $\CC_1$ and $\CC_2$ are orthogonal, the distance between their centers forms the hypotenuse of a right triangle whose legs are equal to the radii.
From Equation of Circle in Cartesian Plane: Formulation 3, the radii $r_1$ and $r_2$ of $\CC_1$ and $\CC_2$ respectively are given by:
\(\ds c_1\) | \(=\) | \(\ds \alpha_1^2 + \beta_1^2 - r_1^2\) | ||||||||||||
\(\ds c_2\) | \(=\) | \(\ds \alpha_2^2 + \beta_2^2 - r_2^2\) |
Hence we have:
\(\ds \paren {\alpha_1 - \alpha_2}^2 + \paren {\beta_1 - \beta_2}^2\) | \(=\) | \(\ds r_1^2 + r_2^2\) | Pythagoras's Theorem and Distance Formula | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha_1^2 + \alpha_2^2 - 2 \alpha_1 \alpha_2 + \beta_1^2 + \beta_2^2 - 2 \beta_1 \beta_2\) | \(=\) | \(\ds r_1^2 + r_2^2\) | multiplying out | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \alpha_1 \alpha_2 + 2 \beta_1 \beta_2\) | \(=\) | \(\ds \paren {\alpha_1^2 + \beta_1^2 - r_1^2} + \paren {\alpha_2^2 + \beta_2^2 - r_2^2}\) | multiplying out | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \alpha_1 \alpha_2 + 2 \beta_1 \beta_2\) | \(=\) | \(\ds c_1 + c_2\) | from above |
$\blacksquare$
Sources
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $16$. condition that two circles should cut orthogonally