Equation of Circle/Parametric

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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:

\(\ds x\) \(=\) \(\ds a + R \cos t\)
\(\ds y\) \(=\) \(\ds b + R \sin t\)

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:

\(\ds \paren {x - a}^2 + \paren {y - b}^2\) \(=\) \(\ds R^2\) Distance Formula
\(\ds \leadsto \ \ \) \(\ds \paren {\dfrac {x - a} R}^2 + \paren {\dfrac {y - b} R}^2\) \(=\) \(\ds 1\) by algebraic manipulations
\(\ds \leadsto \ \ \) \(\ds {x_0}^2 + {y_0}^2\) \(=\) \(\ds 1\) substituting $x_0 = \dfrac {x - a} R$ and $y_0 = \dfrac {y - b} R$

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:

\(\ds \cos \theta\) \(=\) \(\ds \dfrac {x - a} R\)
\(\ds \sin \theta\) \(=\) \(\ds \dfrac {y - b} R\)


If $x - a$ is positive and $y - b$ is negative, it follows from Cosine Function is Even and Sine Function is Odd that:

\(\ds \map \cos {-\theta}\) \(=\) \(\ds \dfrac {x - a} R\)
\(\ds \map \sin {-\theta}\) \(=\) \(\ds \dfrac {y - b} R\)


If $x - a$ is negative and $y - b$ is positive, it follows by Cosine of Supplementary Angle and Sine of Supplementary Angle that:

\(\ds \map \cos {\pi - \theta}\) \(=\) \(\ds \dfrac {x - a} R\)
\(\ds \map \sin {\pi - \theta}\) \(=\) \(\ds \dfrac {y - b} R\)


Finally, if $x - a$ and $y - b$ are both negative:

\(\ds \map \cos {-\pi + \theta}\) \(=\) \(\ds \dfrac {x - a} R\)
\(\ds \map \sin {-\pi + \theta}\) \(=\) \(\ds \dfrac {y - b} R\)


Combining these four cases, we have found $t \in \R$ such that:

\(\ds \cos t\) \(=\) \(\ds \dfrac {x - a} R\)
\(\ds \sin t\) \(=\) \(\ds \dfrac {y - b} R\)


We rearrange these equations to get:

$\begin {cases} x = a + R \cos t \\ y = b + R \sin t \end {cases}$

$\blacksquare$


Sources

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