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Parametric circles and ellipses

Eliminate sine and cosine to identify circles and ellipses. Find centres, radii, semiaxes, restricted arcs, direction and circular arc length using original worked examples.

Before you startEliminating a parameter, sin² θ + cos² θ = 1 and radians

01 / Recognise a circle or ellipse

Scaled sine and cosine satisfy one shared identity.

The rules x = a cos θ and y = b sin θ describe an ellipse when a and b are positive. If a = b, the ellipse is a circle. Adding constants moves the centre; changing signs or swapping sine and cosine changes how the curve is traced.

(x/a)² + (y/b)² = 1

The parameter interval determines whether the whole shape or only an arc is reached. Use the controls to separate shape, direction and permitted arc.

Select an arcExplore

ShapeSign of y termAngle domainTest angle θ

x = 1 + 2 cos θ; y = −1 + 2 sin θ.

(x − 1)² + (y + 1)² = 4.

The whole circle is traced once anticlockwise, starting at (3, −1).

θ = 0 gives (3, −1).

Grey shows the complete shape; blue is the selected arc. Endpoints are included. The parameter θ is generally not the geometric angle of an ellipse’s radius line.

02 / Find the centre and radius

Undo a translation before applying the identity.

Eliminate θ from x = 2 + 3 cos θ and y = −1 + 3 sin θ, for 0 ≤ θ ≤ 2π.Worked example

cos θ = (x − 2)/3; sin θ = (y + 1)/3

Subtract the centre coordinates, then divide by the radius.

(x − 2)²/9 + (y + 1)²/9 = 1

Use cos² θ + sin² θ = 1.

(x − 2)² + (y + 1)² = 9

The circle has centre (2, −1) and radius 3.

Start at (5, −1); trace anticlockwise

At a small positive θ, y increases while x begins to decrease.

The signs inside the Cartesian brackets are opposite to the centre coordinates. A centre with y = −1 produces (y + 1)².

03 / Read semiaxes rather than a single radius

Different horizontal and vertical scales produce an ellipse.

Given x = −2 + 4 cos θ and y = 1 + 2 sin θ over one full turn, identify the locus.Worked example

(x + 2)²/16 + (y − 1)²/4 = 1

Normalise both coordinates and add their squares.

Centre (−2, 1)

Read the translations.

Horizontal semiaxis 4; vertical semiaxis 2

The full width is 8 and full height is 4.

Extremes: (−6, 1), (2, 1), (−2, −1), (−2, 3)

Set sine or cosine to 0 or ±1.

A semiaxis is half an axis length. The denominator 16 corresponds to semiaxis 4, not 16. An ellipse has no single radius that can be substituted into a circle formula.

Watch a circle stretch into an ellipse

Pause, replay or seek freely. The notes explain the same idea and stay in view.

04 / Track direction from the coordinate rules

The same Cartesian equation can hide a different journey.

Compare x = 3 cos θ, y = 2 sin θ with x = 3 sin θ, y = 2 cos θ, as θ increases from 0.Worked example

Both satisfy x²/9 + y²/4 = 1

Their locus is the same complete ellipse over one turn.

The first starts at (3, 0) and moves upwards

It traces anticlockwise.

The second starts at (0, 2) and moves right

It traces clockwise.

Changing y = 2 sin θ to y = −2 sin θ also reverses direction. Squaring during elimination removes that sign, so inspect the original rules whenever direction matters.

05 / Keep the correct restricted arc

A full ellipse equation is too broad for a short parameter interval.

For x = 1 + 3 cos θ and y = −2 + 2 sin θ, let 0 ≤ θ ≤ π.Worked example

(x − 1)²/9 + (y + 2)²/4 = 1

This gives the complete underlying ellipse.

sin θ ≥ 0, so y ≥ −2

Retain only the upper half, relative to its centre.

Start (4, −2), pass through (1, 0), end (−2, −2)

Use θ = 0, π/2 and π.

Both endpoints are included

The parameter inequalities are inclusive.

For 0 ≤ θ ≤ π/2, additionally x ≥ 1: only the upper-right quarter relative to the centre remains. “Upper” refers to the centre, not necessarily to the x-axis.

06 / Check endpoint inclusion and repeated points

An excluded parameter need not imply an excluded point.

For a full standard circle or ellipse, θ = 0 and θ = 2π give the same point. The interval 0 ≤ θ < 2π still covers the whole curve: the starting point is included through θ = 0.

For x = 2 cos θ, y = 2 sin θ, compare three domains.Worked example

0 < θ < π

The upper semicircle excludes both (2, 0) and (−2, 0).

0 ≤ θ < 2π

The whole circle is covered once without repeating the endpoint parameter.

0 < θ < 2π

Only (2, 0) is missing; there is no other parameter for it inside this interval.

When a curve repeats, check all permitted values producing the point before marking a hollow endpoint.

07 / Separate the parameter from the actual angle

A multiplier changes how much curve is traced.

For x = 4 cos(2t) and y = 4 sin(2t), describe 0 ≤ t ≤ π.Worked example

Let φ = 2t

The angle in the coordinate rules runs from 0 to 2π.

x² + y² = 16

The radius is 4.

One full circle is traced anticlockwise

A parameter increase of π gives a geometric angle increase of 2π.

If instead 0 ≤ t ≤ 2π, the circle is traced twice

The locus stays the same, but the travelled distance doubles.

Adding a phase, such as cos(t + π/3) and sin(t + π/3), changes the starting point. It does not move the circle’s centre or change its radius.

08 / Use the actual angular sweep for circular arc length

The formula s = rα uses radians.

For x = 3 cos(2t), y = 3 sin(2t), with π/12 ≤ t ≤ π/3, find the length travelled.Worked example

Actual angle φ = 2t

It runs from π/6 to 2π/3.

Angular sweep = 2π/3 − π/6 = π/2

Use the difference in the actual angle.

Length = 3 × π/2 = 3π/2

The radius is 3, and the angular motion is monotonic.

If the curve retraces, distance travelled counts repeated portions; the length of the distinct locus does not. If the angular rule changes direction, split at the reversals and add absolute angular changes.

For an ellipse, do not use “radius × angle”.
Different semiaxes stretch different directions by different amounts.

09 / Find a parameter from a point

Use both sine and cosine to select the quadrant.

A point on x = 1 + 2 cos θ, y = −1 + 2 sin θ is (0, √3 − 1). Find θ in 0 ≤ θ < 2π.Worked example

cos θ = −1/2; sin θ = √3/2

Subtract the centre and divide by the radius.

θ is in quadrant II

Cosine is negative and sine is positive.

θ = 2π/3

This is the only parameter in the specified single-turn interval.

Using inverse sine alone would also suggest π/3, but that has the wrong x-coordinate. An inverse calculation must satisfy both original coordinate rules.

10 / Write a useful parameterisation from a Cartesian equation

Choose scales, centre and a domain explicitly.

Parameterise (x − 2)²/25 + (y + 1)²/9 = 1, covering only its lower half.Worked example

x = 2 + 5 cos θ; y = −1 + 3 sin θ

The semiaxes are 5 and 3.

π ≤ θ ≤ 2π

Sine is non-positive, selecting y ≤ −1.

A learner says x²/36 + y²/4 = 1 has width 36 and height 4. Correct this.

Hint

First take square roots, then double.

Worked solution

The semiaxes are 6 and 2. The full width is 12 and height is 4; only the stated height happened to be correct.

12 / Recap

Keep shape, arc and direction separate.

Normalise the translated coordinates, then use sin² θ + cos² θ = 1.
Read radius or semiaxes from squared denominators.
Use the parameter interval to select the correct arc and endpoints.
Use the original rules to identify the start and direction.
For circle length, use the actual angular sweep in radians; ellipse length needs a different method.

Section 1 of 12 · Recognise a circle or ellipse