Condition for Circles to be Orthogonal

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

$\blacksquare$