Theorem: Equation of Circle/Parametric

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:
{{begin-eqn}}
{{eqn | l = x
      | r = a + R \cos t
}}
{{eqn | l = y
      | r = b + R \sin t
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \paren {x - a}^2 + \paren {y - b}^2
      | r = R^2
      | c = Distance Formula
}}
{{eqn | ll= \leadsto
      | l = \paren {\dfrac {x - a} R}^2 + \paren {\dfrac {y - b} R}^2
      | r = 1
      | c = by algebraic manipulations
}}
{{eqn | ll= \leadsto
      | l = {x_0}^2 + {y_0}^2
      | r = 1
      | c = substituting $x_0 = \dfrac {x - a} R$ and $y_0 = \dfrac {y - b} R$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \cos \theta
      | r = \dfrac {x - a} R
}}
{{eqn | l = \sin \theta
      | r = \dfrac {y - b} R
}}
{{end-eqn}}
If $x - a$ is positive and $y - b$ is negative, it follows from Cosine Function is Even and Sine Function is Odd that:
{{begin-eqn}}
{{eqn | l = \map \cos {-\theta}
      | r = \dfrac {x - a} R
}}
{{eqn | l = \map \sin {-\theta}
      | r = \dfrac {y - b} R
}}
{{end-eqn}}
If $x - a$ is negative and $y - b$ is positive, it follows by Cosine of Supplementary Angle and Sine of Supplementary Angle that:
{{begin-eqn}}
{{eqn | l = \map \cos {\pi - \theta}
      | r = \dfrac {x - a} R
}}
{{eqn | l = \map \sin {\pi - \theta}
      | r = \dfrac {y - b} R
}}
{{end-eqn}}
Finally, if $x - a$ and $y - b$ are both negative:
{{begin-eqn}}
{{eqn | l = \map \cos {-\pi + \theta}
      | r = \dfrac {x - a} R
}}
{{eqn | l = \map \sin {-\pi + \theta}
      | r = \dfrac {y - b} R
}}
{{end-eqn}}
Combining these four cases, we have found $t \in \R$ such that:
{{begin-eqn}}
{{eqn | l = \cos t
      | r = \dfrac {x - a} R
}}
{{eqn | l = \sin t
      | r = \dfrac {y - b} R
}}
{{end-eqn}}
We rearrange these equations to get:
:$\begin {cases} x = a + R \cos t \\ y = b + R \sin t \end {cases}$
{{qed}}