1CParamCircles


 * Parametric Equations of Circles**

We can use parametric equations to define a circle.
 * x = r cos(t)
 * y = r sin(t)

If we use radians, then t Î [0, 2 p ] .. (or equivalent) If we use degrees, then t Î [0, 360°]

Domain: x Î [–r, r] Range: y Î [–r, r]

Note: if t is set to a domain smaller than 2 p, a partial circle will be obtained.

We can confirm that these parametric equations create a circle by finding the cartesian equation from the equations. This can be done by isolating the trig functions, squaring each equation, then adding the two equations.

math \\ . \qquad x = r \, \cos \big( t \big) \qquad \qquad \qquad y = r \, \sin \big( t \big) \\. \\ . \qquad \cos \big( t \big) = \dfrac{x}{r} \qquad \qquad \qquad \; \sin \big( t \big) = \dfrac{y}{r} math . math \\ . \; \textit{ Square both sides} \\ . \qquad \cos^2 \big( t \big) = \dfrac{x^2}{r^2} \qquad \qquad \quad \;\; \sin^2 \big( t \big) = \dfrac{y^2}{r^2} \\. \\ . \: \textit{ Add the two equations} \\ . \qquad \cos^2 \big( t \big) + \sin^2 \big( t \big) = \dfrac{x^2}{r^2} + \dfrac{y^2}{r^2} math

Using Pythagoras on the triangle shown at the right gives: math . \qquad \cos^2 \big( t \big) + \sin^2 \big( t \big) = 1 math {This result will be revisited in 2C Pythagorean Identities}

Applying this to our equation, we get: math . \qquad \dfrac {x^2}{r^2}+\dfrac{y^2}{r^2}=1 math

Multiply both sides by r 2 gives math . \qquad x^2 + y^2 = r^2 math

This is the equation of a circle with its centre at the origin and a radius of r.

General Parametric Equations of Circles

For a circle that has been translated h units to the right and k units up, the parametric equations become:
 * x = r cos(t) + h
 * y = r sin(t) + k

Center at (h, k)

If we use radians, then t Î [0, 2 p ] (or equivalent) If we use degrees, then t Î [0, 360°]

Domain: x Î [h – r, h + r] Range: y Î [k – r, k + r]

Example

Find the cartesian equation of the circle with parametric equations
 * x = 2 cos(t) + 1 ... ... and
 * y = 2 sin(t) + 3 ... ... t Î [0, 2 p ]

This can be done by isolating the trig functions, squaring each equation, then adding the two equations.

math \\ . \qquad x = 2 \cos \big( t \big) + 1 \qquad \qquad \qquad y = 2 \sin \big( t \big) + 3 \\. \\ . \qquad 2 \cos \big( t \big) = x - 1 \qquad \qquad \qquad 2 \sin \big( t \big) = y - 3 math . math \\ . \qquad \cos \big( t \big) = \dfrac{x-1}{2} \qquad \qquad \qquad \sin \big( t \big) = \dfrac{y-3}{2} \\. \\ . \; \textit{ Square both sides} \\ . \qquad \cos^2 \big( t \big) = \dfrac{ (x-1)^2}{4} \qquad \qquad \qquad \sin^2 \big( t \big) = \dfrac{(y-3)^2}{4} math . math \\ . \; \textit{ Add the two equations} \\ . \qquad \cos^2 \big( t \big) + \sin^2 \big( t \big) = \dfrac{ (x-1)^2}{4} + \dfrac{(y-3)^2}{4} math . math \\ . \; \textit{ Using Pythagoras} \\ . \qquad \cos^2(t)+\sin^2(t)=1 math . math . \qquad \dfrac{(x-1)^2}{4}+\dfrac{(y-3)^2}{4}=1 \qquad \big\{ \textit{Multiply both sides by } 4 \big\} \\. \\ . \qquad (x-1)^2+(y-3)^2=4 math

This is the cartesian equation of a circle with centre at (1, 3) and radius = 2

math \\ . \qquad \text {Domain: } x \in \lbrack 1-2, \, 1+2 \rbrack \Longrightarrow x \in \lbrack -1, \, 3 \rbrack \\. \\ . \qquad \text {Range: } \;\; y \in \lbrack 3-2, \, 3+2 \rbrack \Longrightarrow y \in \lbrack 1, \, 5 \rbrack math

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