# Equation of the Union of Two Circles

$(x^{2}+y^{2})^{2}=36+13(x^{2}+y^{2})$ is an equation of a pair of concentric circles.

A circle of radius $r$ centered at the origin is described by the equation $x^{2}+y^{2}-r^{2}=0.$ This is a particular case of a general second order, or *quadratic*, equation. The equation above is of order $4,$ (it's a so-called *quartic equation*), because $4$ is the highest order of the terms in the equation. Indeed, there are three such terms: $x^{4},$ $y^{4},$ and $x^{2}y^{2}.$

Assume $f(x,y)=0$ describes a set of points (a curve, perhaps) in the plane, call it $A$; let $g(x,y)=0$ describe another set of points, say $B.$. Then $f(x,y)g(x,y)=0$ describes the union of the two sets. Symbolically:

$A\cup B=\{(x,y):\space f(x,y)g(x,y)=0\}.$

Why is that so?

$\Rightarrow$

If $(x,y)\in A\cup B,$ i.e., if point $(x,y)$ belongs to the union of the two sets, it belongs to at least one of them. Without loss of generality, we may assume that this is set $A:$ $(x,y)\in A.$ This exactly means that $(x,y)$ satisfies the equation $f(x,y)=0.$ But then it is also true that $f(x,y)g(x,y)=0.$

$\Leftarrow$

Assume that, for a point $(x,y),$ $f(x,y)g(x,y)=0.$ But, if a product of two factors (in the field of real or complex numbers) is zero, then at least one of the factors must be zero. Without loss of generality, we may assume that $f(x,y)=0,$ implying that $(x,y)\in A\subset A\cup B,$ and we are done.

Now, let's return to the equation at the top of the page: $(x^{2}+y^{2})^{2}=36+13(x^{2}+y^{2}).$ If there is not mistake, this equation could be rewritten as

$(x^{2}+y^{2}-4)(x^{2}+y^{2}-9)=0.$

The equation is factored into two: $x^{2}+y^{2}-4=0$ and $x^{2}+y^{2}-9=0,$ each describing a circle centered at the origin.

It is really easy to come up with an equation of the union of two curves which, after the "simplification" procedure - collecting the like terms - may become next to inscrutable. For example, whose equation is this: $x^{4}+2x^{2}y^{2}+y^{4}-15x^{2}=0?$

What equation describes the intersection of two sets? The answer may depend on one's outlook. One is that the intersection is described by a system of equations:

$(x,y)\in A\cap B=\{(x,y):\space \begin{cases}f(x,y)=0\\g(x,y)=0\end{cases}\space\}. $

Another possibility is to introduce a vector function $\mathbf{F}:$ $\mathbf{F}(x,y)=(f(x,y),g(x,y))$ such that the intersection of the two sets will be described by a single vector equation:

$(x,y)\in A\cap B=\{(x,y):\space \mathbf{F}(x,y)=0\}. $

As I was reminded by Andrey Mokhov, a vector is $\mathbf{0}$ if its norm is $0.$ Taking the Euclidean norm shows that the intersection of two sets can also be described by a single equation:

$(x,y)\in A\cap B=\{(x,y):\space ||\mathbf{F}(x,y)||^2=f^{2}(x,y)+g^{2}(x,y)=0\}. $

Any other norm will do as well. For example,

$(x,y)\in A\cap B=\{(x,y):\space ||\mathbf{F}(x,y)||_{1}=|f(x,y)|+|g(x,y)|=0\}, $

or,

$(x,y)\in A\cap B=\{(x,y):\space ||\mathbf{F}(x,y)||_{\infty}=\max\{|f(x,y)|,|g(x,y)|\}=0\}. $

### Reference

- K. Kendig,
*Conics*, MAA, 2005

|Contact| |Front page| |Contents| |Did you know?| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny63602191 |