nyuuzyou/wikis huggingface dataset:
Equation of Catenary/Cesàro/Formulation 2

== Theorem ==
The catenary can be described by the Cesàro equation:
:$a \rho = a^2 + s^2$
where:
:$s$ is the arc length
:$\rho$ is the radius of curvature
:$a$ is a constant.
== Proof ==
== Historical Note ==
== Lingustic Note ==
== Sources ==
* 
* 
* 
Category:Cesàro Equation for Catenary
Category:Catenary
Category:Cesàro Equations
Equation of Catenary/Whewell

== Theorem ==
The catenary can be described by the Whewell equation:
:$s = a \tan \psi$
where:
:$s$ is the arc length
:$\psi$ is the turning angle
:$a$ is a constant.
== Proof ==
By definition, the catenary is the shape made by an ideally flexible chain hanging freely from two arbitrary points.
Let the catenary $\CC$ lie in a Cartesian plane.
Let the lowest point on $\CC$ be $P_0$.
Let $P$ be an arbitrary point on the chain.
Let $s$ be the length along the arc of the chain from the $P_0$ to $P$.
Let $w$ be the linear mass density of the chain, that is, its weight per unit length.
The section of chain between the lowest point and $P$ is in static equilibrium under the influence of three forces, as follows:
:The tension $T_0$ at $P_0$
:The tension $T$ at $P$
:The weight $w s$ of the chain between these two points.
As the chain is (ideally) flexible, the tension $T$ is along the line of the chain, and therefore along a tangent to the chain.
At $P_0$, the tangent to $\CC$ is a horizontal line.
Let the angle between the tangent to $\CC$ at $P$ and that (horizontal) tangent to $\CC$ at $P_0$ be $\psi$.
By definition, $\psi$ is the turning angle of $\CC$.
:
Let us resolve $T$ into its horizontal and vertical components:
:$T_0 = T \cos \psi$
:$w s = T \sin \psi$
We divide one by the other to eliminate $T$ and set $a = T_0 / w$:
:$\tan \psi = \dfrac s a$
from which the result follows.
== Historical Note ==
== Lingustic Note ==
== Sources ==
* 
* 
Category:Catenary
Category:Whewell Equations
Equation of Chord of Contact on Circle Centered at Origin

== Theorem ==
Let $\CC$ be a circle whose radius is $r$ and whose center is at the origin of a Cartesian plane.
Let $P = \tuple {x_0, y_0}$ be a point which is outside the boundary of $\CC$.
Let $UV$ be the chord of contact on $\CC$ with respect to $P$.
Then $UV$ can be defined by the equation:
:$x x_0 + y y_0 = r^2$
== Proof ==
Let $\TT_1$ and $\TT_2$ be a tangents to $\CC$ passing through $P$.
Let:
:$\TT_1$ touch $\CC$ at $U = \tuple {x_1, y_1}$
:$\TT_2$ touch $\CC$ at $V = \tuple {x_2, y_2}$
Then the chord of contact on $\CC$ with respect to $P$ is defined as $UV$.
:
From Equation of Tangent to Circle Centered at Origin, $\TT_1$ is expressed by the equation:
:$x x_1 + y y_1 = r^2$
but as $\TT_1$ also passes through $\tuple {x_0, y_0}$ we also have:
:$x_0 x_1 + y_0 y_1 = r^2$
This also expresses the condition that $U$ should lie on $\TT_1$:
:$x x_0 + y y_0 = r^2$
Similarly, From Equation of Tangent to Circle Centered at Origin, $\TT_2$ is expressed by the equation:
:$x x_2 + y y_2 = r^2$
but as $\TT_2$ also passes through $\tuple {x_0, y_0}$ we also have:
:$x_0 x_2 + y_0 y_2 = r^2$
This also expresses the condition that $V$ should lie on $\TT_2$:
:$x x_0 + y y_0 = r^2$
So both $U$ and $V$ lie on the straight line whose equation is:
:$x x_0 + y y_0 = r^2$
and the result follows.
== Sources ==
* : Chapter $\text {III}$. The Circle: $4$. Pole and polar
Category:Circles
Category:Chords of Contact
Equation of Chord of Contact on Ellipse in Reduced Form

== Theorem ==
Let $\EE$ be an ellipse embedded in a Cartesian plane in reduced form with the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Let $P = \tuple {x_0, y_0}$ be a point which is outside the boundary of $\EE$.
Let $UV$ be the chord of contact on $\EE$ with respect to $P$.
Then $UV$ can be defined by the equation:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$
== Proof ==
Let $\TT_1$ and $\TT_2$ be a tangents to $\EE$ passing through $P$.
Let:
:$\TT_1$ touch $\EE$ at $U = \tuple {x_1, y_1}$
:$\TT_2$ touch $\EE$ at $V = \tuple {x_2, y_2}$
Then the chord of contact on $\EE$ with respect to $P$ is defined as $UV$.
:
From Equation of Tangent to Ellipse in Reduced Form, $\TT_1$ is expressed by the equation:
:$\dfrac {x x_1} {a^2} + \dfrac {y y_1} {b^2} = 1$
but as $\TT_1$ also passes through $\tuple {x_0, y_0}$ we also have:
:$\dfrac {x_0 x_1} {a^2} + \dfrac {y_0 y_1} {b^2} = 1$
This also expresses the condition that $U$ should lie on $\TT_1$:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$
Similarly, From Equation of Tangent to Circle Centered at Origin, $\TT_2$ is expressed by the equation:
:$\dfrac {x x_2} {a^2} + \dfrac {y y_2} {b^2} = 1$
but as $\TT_2$ also passes through $\tuple {x_0, y_0}$ we also have:
:$\dfrac {x_0 x_2} {a^2} + \dfrac {y_0 y_2} {b^2} = 1$
This also expresses the condition that $V$ should lie on $\TT_2$:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$
So both $U$ and $V$ lie on the straight line whose equation is:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$
and the result follows.
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $3$.
Category:Ellipses
Category:Chords of Contact
Equation of Circle

== Theorem ==
=== Cartesian Coordinates ===
=== Parametric Equation ===
=== Polar Coordinates ===
In polar coordinates, it does not make sense to refer to a point by $x$ and $y$ coordinates.
Instead, the center of a circle is commonly denoted $\polar {r_0, \varphi}$, where $r_0$ is the distance from the origin and $\varphi$ is the angle from the polar axis in the counterclockwise direction.
The equation for a circle with radius $R$ of this type is:
Category:Circles
Equation of Circle/Cartesian/Formulation 1

== Theorem ==
The equation of a circle embedded in the Cartesian plane with radius $R$ and center $\tuple {a, b}$ can be expressed as:
:$\paren {x - a}^2 + \paren {y - b}^2 = R^2$
== Proof ==
Let the point $\tuple {x, y}$ satisfy the equation:
:$(1): \quad \paren {x - a}^2 + \paren {y - b}^2 = R^2$
By the Distance Formula, the distance between this $\tuple {x, y}$ and $\tuple {a, b}$ is:
:$\sqrt {\paren {x - a}^2 + \paren {y - b}^2}$
But from equation $(1)$, this quantity equals $R$.
Therefore the distance between points satisfying the equation and the center is constant and equal to the radius.
Thus $\tuple {x, y}$ lies on the circumference of a circle with radius $R$ and center $\tuple {a, b}$.
Now suppose that $\tuple {x, y}$ does not satisfy the equation:
:$\paren {x - a}^2 + \paren {y - b}^2 = R^2$
Then by the same reasoning as above, the distance between $\tuple {x, y}$ and $\tuple {a, b}$ does not equal $R$.
Therefore $\tuple {x, y}$ does not lie on the circumference of a circle with radius $R$ and center $\tuple {a, b}$.
Hence it follows that the points satisfying $(1)$ are exactly those points which are the circle in question.
== Sources ==
* : Chapter $\text {III}$. The Circle: $14$. To find the equation of the circle whose centre is $\tuple {\alpha, \beta}$ and radius $r$
* : $\S 10$: Formulas from Plane Analytic Geometry: $10.15$: Equation of Circle of Radius $R$, Center are $\tuple {x_0, y_0}$
* 
* 
* 
* 
Category:Equation of Circle
Equation of Circle/Cartesian/Formulation 2

== Theorem ==
The equation:
:$A \paren {x^2 + y^2} + B x + C y + D = 0$
is the equation of a circle with radius $R$ and center $\tuple {a, b}$, where:
:$R = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}$
:$\tuple {a, b} = \tuple {\dfrac {-B} {2 A}, \dfrac {-C} {2 A} }$
provided:
:$A > 0$
:$B^2 + C^2 \ge 4 A D$
== Proof ==
This last expression is non-negative $B^2 + C^2 \ge 4 A D$.
In such a case $\dfrac 1 {4 A^2} \paren {B^2 + C^2 - 4 A D}$ is in the form $R^2$ and so:
:$\paren {x + \dfrac B {2 A} }^2 + \paren {y + \dfrac C {2 A} }^2 = \dfrac 1 {4 A^2} \paren {B^2 + C^2 - 4 A D}$
is in the form:
:$\paren {x - a}^2 + \paren {y - b}^2 = R^2$
Hence the result from Equation of Circle in Cartesian Plane: Formulation 1.
== Sources ==
* 
* 
Category:Equation of Circle
Equation of Circle/Cartesian/Formulation 3

== Equation of Circle in Cartesian Plane: Also presented as ==
The equation of a circle with radius $R$ and center $\tuple {a, b}$ embedded in the Cartesian plane can be expressed as:
:$x^2 + y^2 - 2 a x - 2 b y + c = 0$
where:
:$c = a^2 + b^2 - R^2$ 
== Proof ==
== Sources ==
* : Chapter $\text {III}$. The Circle: $14$. To find the equation of the circle whose centre is $\tuple {\alpha, \beta}$ and radius $r$
Category:Equation of Circle
Equation of Circle/Parametric

== Theorem ==
The equation of a circle embedded in the Cartesian plane with radius $R$ and center $\tuple {a, b}$ can be expressed as a parametric equation:
:$\begin {cases} x = a + R \cos t \\ y = b + R \sin t \end {cases}$
== Proof ==
Let the point $\tuple {x, y}$ satisfy the equations:
By the Distance Formula, the distance between $\tuple {x, y}$ and $\tuple {a, b}$ is:
:$\sqrt {\paren {\paren {a + R \cos t} - a}^2 + \paren {\paren {b + R \sin t} - b}^2}$
This simplifies to:
:$\sqrt {R^2 \cos^2 t + R^2 \sin^2 t} = R \sqrt {\cos^2 t + \sin^2 t}$
Then by Sum of Squares of Sine and Cosine, this distance equals $R$.
Therefore the distance between points satisfying the equation and the center is constant and equal to the radius.
Let the point $\tuple {x, y}$ lie on the embedded circle with radius $R$ and center $\tuple {a, b}$.
As the distance between $\tuple {x, y}$ and $\tuple {a, b}$ is equal to $R$, it follows that:
where $\size {x_0}, \size {y_0} \in \closedint 0 1$, otherwise we would have ${x_0}^2 + {y_0}^2 > 1$.
From Cosine of Integer Multiple of Pi:
:$\cos 0 = 1$
From Zeroes of Cosine:
:$\map \cos {\dfrac \pi 2} = 0$
By Cosine Function is Continuous and the Intermediate Value Theorem:
:$\exists \theta \in \closedint 0 {\dfrac \pi 2}: \cos \theta = \size {x_0}$
From Sum of Squares of Sine and Cosine:
:$\cos^2 \theta + \sin^2 \theta = 1$
Hence:
:$\sin \theta = \size {y_0}$
If $x - a$ and $y - b$ are both positive:
If $x - a$ is positive and $y - b$ is negative, it follows from Cosine Function is Even and Sine Function is Odd that:
If $x - a$ is negative and $y - b$ is positive, it follows by Cosine of Supplementary Angle and Sine of Supplementary Angle that:
Finally, if $x - a$ and $y - b$ are both negative:
Combining these four cases, we have found $t \in \R$ such that:
We rearrange these equations to get:
:$\begin {cases} x = a + R \cos t \\ y = b + R \sin t \end {cases}$
== Sources ==
* : Chapter $\text{I}$: The Theory of Space Curves: $1$. Introductory remarks about space curves
* 
* 
Category:Equation of Circle
Equation of Circle/Polar

== Theorem ==
The equation of a circle with radius $R$ and center $\polar {r_0, \varphi}$ can be expressed in polar coordinates as:
:$r^2 - 2 r r_0 \map \cos {\theta - \varphi} + \paren {r_0}^2 = R^2$
where:
:$r_0$ is the distance of the center from the origin
:$\varphi$ is the angle of the center from the polar axis in the counterclockwise direction
:$r$ is a function of $\theta$.
=== Corollary ===
== Proof ==
Let the point $\polar {r, \theta}_\text {Polar}$ satisfy the equation:
:$r^2 - 2 r r_0 \map \cos {\theta - \varphi} + \paren {r_0}^2 = R^2$
Let the points $\polar {r, \theta}$ and $\polar {r_0, \varphi}$ be rewritten in Cartesian coordinates:
:$\polar {r, \theta}_\text {Polar} = \tuple {r \cos \theta, r \sin \theta}_\text{Cartesian}$
:$\polar {r_0, \varphi}_\text{Polar} = \tuple {r_0 \cos \varphi, r_0 \sin \varphi}_\text{Cartesian}$
Thus the distance between $\polar {r, \theta}_\text {Polar}$ and $\polar {r_0, \varphi}_\text{Polar}$ is:
But from the equation, this quantity equals $R$.
Therefore the distance between points satisfying the equation and the center is constant and equal to the radius $R$.
Category:Equation of Circle
Equation of Circle/Polar/Corollary

== Corollary to Equation of Circle ==
Let $C$ be a circle whose radius is $R$.
Let $C$ be aligned in a polar coordinate frame such that its center is at the origin.
Then the equation of a $C$ is given by:
: $r = R$
== Proof ==
From Equation of Circle: Polar Form, we have a circle whose center is at $\polar {r_0, \varphi}$ whose radius is $R$ is:
:$r^2 - 2 r r_0 \, \map \cos {\theta - \varphi} + \paren {r_0}^2 = R^2$
So, when $\polar {r_0, \varphi} = \polar {0, 0}$:
Category:Equation of Circle
Equation of Circle center Origin

== Theorem ==
The circle with radius $R$ whose center is at the origin expressed in Cartesian coordinates is given by the equation:
:$x^2 + y^2 = R^2$
== Proof ==
From Equation of Circle in Cartesian Plane, the equation of a circle with radius $R$ and center $\tuple {a, b}$ expressed in Cartesian coordinates is:
:$\paren {x - a}^2 + \paren {y - b}^2 = R^2$
Setting $a = b = 0$ yields the result.
== Sources ==
* : Chapter $\text I$: Functions and their Graphs: $1$. Constants and Variables
* : Chapter $\text {III}$. The Circle: $1$. To find the equation of a circle of radius $r$ with its centre at the origin
* : Chapter $\text{III}$: Gentleman, Soldier and Mathematician
* : Chapter $\text{I}$: The Theory of Space Curves: $1$. Introductory remarks about space curves
* : Chapter $\text {B}.6$: How Archimedes Discovered Integration
* 
* 
* : Chapter $6$: Curves and Coordinates: Descartes
Category:Equation of Circle
Equation of Circle in Complex Plane/Examples

== Examples of Use of Equation of Circle in Complex Plane ==
=== Radius $4$, Center $\tuple {-2, 1}$ ===
=== Radius $2$, Center $\tuple {0, 1}$ ===
=== Radius $2$, Center $\tuple {-3, 4}$ ===
=== Radius $4$, Center $\tuple {0, -3}$ ===
=== Annulus: $1 < \cmod {z + i} \le 2$ ===
=== Circle Defined by $z \paren {\overline z + 2} = 3$ ===
=== Radius $4$, Center $\tuple {0, 0}$ ===
=== Imaginary Radius $2$, Center $\tuple {2, 0}$ ===
=== Degenerate Case: Straight Line $x = 2$ ===
=== Degenerate Case: Straight Line $y = -3$ ===
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/1 less than mod (z + i) not greater than 2

== Example of Use of Equation of Circle in Complex Plane ==
The inequality:
:$1 < \cmod {z + i} \le 2$
describes the inside of the annulus whose center is at $-i$, whose inner radius is $1$ and whose outer radius is $2$.
This annulus does not include its inner boundary, but does include its outer boundary.
:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $73 \ \text {(a)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/Imaginary Radius 2, Center (2, 0)

== Example of Use of Equation of Circle in Complex Plane ==
The equation:
:$z \overline z - 2 z - 2 \overline z + 8 = 0$
describes a circle embedded in the complex plane whose center is at $\tuple {2, 0}$ and whose radius is $2$ imaginary units.
== Proof ==
The result follows from Equation of Circle in Cartesian Plane, except that the square of the radius is $-4$.
Hence the radius has do be described as $2$ imaginary units.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Conjugate Coordinates: $116 \ \text{(b)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/Radius 2, Center (-3, 4)

== Example of Use of Equation of Circle in Complex Plane ==
Let $C$ be a circle embedded in the complex plane whose radius is $2$ and whose center is $\paren {-3, 4}$.
Then $C$ can be described by the equation:
:$\cmod {z + 3 - 4 i} = 2$
or in conventional Cartesian coordinates:
:$\paren {x + 3}^2 + \paren {y - 4}^2 = 4$
== Proof ==
From Equation of Circle in Complex Plane, a circle whose radius is $r$ and whose center is $\alpha$ has equation:
:$\cmod {z - \alpha} = r$
Substituting $\alpha = -3 + 4 i$ and $r = 2$ gives:
:$\cmod {z - \paren {-3 + 4 i} } = 2$
that is:
:$\cmod {z + 3 - 4 i} = 2$
Letting $z = x + i y$ gives:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $72 \ \text {(a)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/Radius 2, Center (0, 1)

== Example of Use of Equation of Circle in Complex Plane ==
Let $C$ be a circle embedded in the complex plane whose radius is $2$ and whose center is $\paren {0, 1}$.
Then $C$ can be described by the equation:
:$\cmod {z - i} = 2$
or in conventional Cartesian coordinates:
:$x^2 + \paren {y - 1}^2 = 4$
== Proof ==
From Equation of Circle in Complex Plane, a circle whose radius is $r$ and whose center is $\alpha$ has equation:
:$\cmod {z - \alpha} = r$
:
Substituting $\alpha = i$ and $r = 2$ gives:
:$\cmod {z - i} = 2$
Letting $z = x + i y$ gives:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $71 \ \text {(a)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/Radius 4, Center (-2, 1)

== Example of Use of Equation of Circle in Complex Plane ==
Let $C$ be a circle embedded in the complex plane whose radius is $4$ and whose center is $\paren {-2, 1}$.
Then $C$ can be described by the equation:
:$\cmod {z + 2 - i} = 4$
or in conventional Cartesian coordinates:
:$\paren {x + 2}^2 + \paren {y - 1}^2 = 16$
== Proof ==
From Equation of Circle in Complex Plane, a circle whose radius is $r$ and whose center is $\alpha$ has equation:
:$\cmod {z - \alpha} = r$
:
Substituting $\alpha = -2 + i$ and $r = 4$ gives:
:$\cmod {z + 2 - i} = 4$
Letting $z = x + i y$ gives:
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $13 \ \text{(a)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/Radius 4, Center (0, -3)

== Example of Use of Equation of Circle in Complex Plane ==
The inequality:
:$\cmod {z + 3 i} > 4$
describes the area outside the circle whose center is at $-3 i$, whose radius is $4$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $73 \ \text {(c)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/Radius 4, Center (0, 0)

== Example of Use of Equation of Circle in Complex Plane ==
The equation:
:$z \overline z = 16$
describes a circle embedded in the complex plane whose center is at $\tuple {0, 0}$ and whose radius is $4$.
== Proof ==
The result follows from Equation of Circle in Cartesian Plane.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Conjugate Coordinates: $116 \ \text{(a)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/Straight Line x = 2

== Example of Use of Equation of Circle in Complex Plane ==
The equation:
:$z + \overline z = 4$
describes the straight line $x = 2$ embedded in the complex plane 
== Proof ==
This is an instance of the Equation of Circle in Complex Plane: Formulation 2:
:$\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$
where $\alpha = 0$.
This is a straight line $\alpha = 0$ and $\beta \ne 0$.
Hence:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Conjugate Coordinates: $116 \ \text{(c)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/Straight Line y = -3

== Example of Use of Equation of Circle in Complex Plane ==
The equation:
:$\overline z = z + 6 i$
describes the straight line $y = -3$ embedded in the complex plane 
== Proof ==
This is an instance of the Equation of Circle in Complex Plane: Formulation 2:
:$\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$
where $\alpha = 0$.
This is a straight line $\alpha = 0$ and $\beta \ne 0$.
Hence:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Conjugate Coordinates: $116 \ \text{(d)}$
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Examples/z (conj z + 2) = 3

== Example of Use of Equation of Circle in Complex Plane ==
The equation:
:$z \paren {\overline z + 2} = 3$
is a quadratic equation with $2$ solutions:
:$z = 1$
:$z = -3$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $71 \ \text {(d)}$
Category:Quadratic Equations
Equation of Circle in Complex Plane/Examples/z (conj z + 2) = 3/Mistake

== Source Work ==
:Chapter $1$: Complex Numbers
::Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $71 \ \text {(d)}$
== Mistake ==
:Describe and graph the locus represented by each of the following:
::... $\text (d)$ $z \paren {\overline z + 2} = 3$
:Ans. ... $\text (d)$ circle, ...
== Correction ==
Working through in the direction one would go when trying to demonstrate the locus is a circle:
While it looks like the result follows from Equation of Circle in Cartesian Plane:
:
it does not, because $y$ is real.
Otherwise, the circle would have center $\tuple {-1, -i} \notin \R^2$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $71 \ \text {(d)}$
Category:Mistakes/Theory and Problems of Complex Variables
Equation of Circle in Complex Plane/Formulation 1

== Theorem ==
Let $\C$ be the complex plane.
Let $C$ be a circle in $\C$ whose radius is $r \in \R_{>0}$ and whose center is $\alpha \in \C$.
Then $C$ may be written as:
:$\cmod {z - \alpha} = r$
where $\cmod {\, \cdot \,}$ denotes complex modulus.
=== Interior ===
=== Exterior ===
== Proof ==
Let $z = x + i y$.
Let $\alpha = a + i b$.
Thus:
The result follows from Equation of Circle.
== Sources ==
* : $\S 2$. Geometrical Representations: Example $1$.
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Formulation 1/Exterior

== Theorem ==
Let $\C$ be the complex plane.
Let $C$ be a circle in $\C$ whose radius is $r \in \R_{>0}$ and whose center is $\alpha \in \C$.
The points in $\C$ which correspond to the exterior of $C$ can be defined by:
:$\left\lvert{z - \alpha}\right\rvert > r$
where $\left\lvert{\, \cdot \,}\right\rvert$ denotes complex modulus.
== Proof ==
From Equation of Circle in Complex Plane, the circle $C$ itself is given by:
:$\left\lvert{z - \alpha}\right\rvert = r$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $1$.
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Formulation 1/Interior

== Theorem ==
Let $\C$ be the complex plane.
Let $C$ be a circle in $\C$ whose radius is $r \in \R_{>0}$ and whose center is $\alpha \in \C$.
The points in $\C$ which correspond to the interior of $C$ can be defined by:
:$\cmod {z - \alpha} < r$
where $\cmod {\, \cdot \,}$ denotes complex modulus.
== Proof ==
From Equation of Circle in Complex Plane, the circle $C$ itself is given by:
:$\cmod {z - \alpha} = r$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $1$.
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Formulation 2

== Theorem ==
Let $\C$ be the complex plane.
Let $C$ be a circle in $\C$.
Then $C$ may be written as:
:$\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$
where:
:$\alpha \in \R_{\ne 0}$ is real and non-zero
:$\gamma \in \R$ is real
:$\beta \in \C$ is complex such that $\cmod \beta^2 > \alpha \gamma$.
The curve $C$ is a straight line $\alpha = 0$ and $\beta \ne 0$.
== Proof 1 ==
== Proof 2 ==
Category:Equation of Circle in Complex Plane
Equation of Circle in Complex Plane/Formulation 2/Proof 1

== Theorem ==
== Proof ==
By setting:
: $\beta := -\alpha a$ and $\gamma = \alpha a \overline a - r^2$
we have:
:$\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$
We have that:
If $\alpha = 0$ and $\beta \ne 0$ the equation devolves to:
:$\beta z + \overline \beta \overline z + \gamma = 0$
which from Equation of Line in Complex Plane: Formulation 1 is the equation of a straight line.
The result follows.
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $5$.
Category:Equation of Circle in Complex Plane