Circles — Area, Circumference, Sectors

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Key Circle Vocabulary:

• Radius (r): distance from centre to any point on the circle
• Diameter (d): 2r, passes through the centre
• Circumference: the perimeter = 2πr = πd
• Area: πr²

Sector of a circle:

• Arc length = (θ/360°) × 2πr (where θ is the central angle in degrees)
• Sector area = (θ/360°) × πr²
• If using radians: arc length = rθ, sector area = ½r²θ

Annulus (ring between two concentric circles):

• Area = π(R² − r²), where R is the outer radius and r is the inner radius

⚡ Exam tip: When asked for “the area of the shaded region,” look for symmetry and complementary areas. Often the shaded region is the sector minus a triangle or circle.

⚡ Exam tip: π ≈ 22/7 is useful for fraction-based questions; π ≈ 3.14 for decimal approximations. If the answer is left in terms of π, that’s usually the expected form.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Arc Length and Sector Area — Working with Angles

The fraction of the circle a sector occupies equals θ/360°, where θ is the central angle. A 90° sector of a circle with radius 8 cm has area = (90/360) × π × 8² = (¼) × 64π = 16π cm² and arc length = (90/360) × 2π × 8 = 4π cm.

For a semicircle (θ = 180°), area = ½πr² and arc length = πr (the curved part, not including diameter).

Chords and the Perpendicular Bisector

A chord’s perpendicular bisector passes through the circle’s centre. The distance d from the centre to a chord of length c is: d = √(r² − (c/2)²). A radius drawn to a chord’s midpoint is perpendicular to the chord.

If a circle has radius 10 cm and a chord is 12 cm long, the distance from centre to chord = √(100 − 36) = √64 = 8 cm.

Tangent Properties

A tangent to a circle is perpendicular to the radius at the point of contact. Two tangents drawn from an external point to a circle are equal in length. If point P is outside a circle with centre O, and tangents PA and PB touch the circle at A and B, then PA = PB.

Segment of a Circle

A segment is the region between a chord and its arc. Area of segment = sector area − triangle area (where the triangle has the chord as its base and the centre as its vertex).

For a minor segment with chord length c subtending angle θ at centre:

• Triangle area = ½r² sin θ
• Segment area = ½r²(θ − sin θ) in radians, or using degrees: = (πr²θ/360) − (½r² sin(θ))

Common Mistakes to Avoid:

Mistake	Correct approach
Confusing arc length with sector area	Arc length is a length (units: cm); sector area is an area (units: cm²)
Forgetting to convert angle to radians when required	Most circle formulae use degrees unless specified; check the formula
Not drawing a radius to the point of tangency when solving tangent problems	The radius-to-tangent point is always perpendicular to the tangent
Using diameter instead of radius in sector area formula	Area = πr², NOT πd²/4 (though they’re mathematically equivalent, r² is simpler)

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

The Equation of a Circle

In Cartesian coordinates, a circle with centre (a, b) and radius r has equation: (x − a)² + (y − b)² = r². The unit circle (centre at origin, radius 1) is x² + y² = 1.

From the general form x² + y² + 2gx + 2fy + c = 0, complete the square to find centre (−g, −f) and radius √(g² + f² − c).

Example: x² + y² − 6x + 8y + 9 = 0 → (x² − 6x) + (y² + 8y) = −9 → (x − 3)² − 9 + (y + 4)² − 16 = −9 → (x − 3)² + (y + 4)² = 16. Centre (3, −4), radius = 4.

Angles in a Circle — Theorems

The angle subtended by a chord at the circumference is half the angle subtended at the centre. If chord AB subtends 40° at the centre O, it subtends 20° at any point on the circumference.

Angles in the same segment (on the same side of a chord) are equal. Angles in a semicircle are 90° (Thales’ theorem).

Cyclic quadrilaterals: opposite angles sum to 180°. If angle A = 70° and angle C = 110°, they’re supplementary. An exterior angle of a cyclic quadrilateral equals the opposite interior angle.

Two-Dimensional Packing Problems

The UI exam occasionally tests how many circles of a given radius fit inside a larger circle, or how many small circles fit in a sector. The approach is usually estimation based on area ratios, though exact answers may require geometric reasoning.

Mixed Area Problems

These are common and require decomposing the figure into standard shapes. Consider a figure consisting of a rectangle 10cm × 6cm with a semicircle of diameter 10cm attached to one short side. Total area = rectangle area + semicircle area = 60 + ½π(5)² = 60 + 12.5π ≈ 99.27 cm².

Compound Shapes Involving Circles

The classic “四个圆互相外切” type: four circles of radius r are arranged so each touches two neighbours. The centres form a square of side 2r. The region in the centre (curvilinear quadrilateral) can be found by adding four sector areas and subtracting the square.

For circles inscribed in sectors: if a circle of radius r is inscribed in a sector of radius R and angle θ (radians), the relationship is r = R sin(θ/2) / (1 + sin(θ/2)).

Historical Context

The ratio circumference/diameter (π) has been studied since antiquity. Archimedes (c. 250 BCE) bounded π between 3 10/71 and 3 1/7 using inscribed and circumscribed polygons. The Egyptian Rhind Papyrus (c. 1650 BCE) gives π ≈ 3.16. The symbol π was first used by William Jones in 1706 and popularised by Leonhard Euler in 1737.

Exam Pattern Analysis

UI circle questions commonly appear as:

1. Calculate arc length, sector area, or segment area given radius and angle
2. Find area of shaded region in composite figures (often involving quarter circles or semicircles)
3. Apply tangent properties to find missing lengths
4. Use angle theorems (subtended angles, cyclic quadrilaterals)
5. Solve problems combining circles with other shapes

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