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