Circles — centre, radius, diameter, chord, arc, sector, segment

Overview

A circle is the set of all points in a plane that are equidistant from a fixed point. Understanding its parts is fundamental to solving geometric problems involving circular shapes, from wheel design to planetary orbits.

Core Definitions

Derivation from first principles:

• A diameter connects two points on the circle, passing through centre $O$
• Let's call these points $A$ and $B$, with $O$ between them
• By definition of radius: $AO = r$ and $OB = r$
• Since $A$, $O$, $B$ are collinear (on same line): $AB = AO + OB = r + r = 2r$
• Why this step? We're using the segment addition property: when three points lie on a straight line, the distance from first to third equals the sum of distances first-to-middle and middle-to-third.

Key property: The perpendicular from the centre to a chord bisects the chord.

Derivation of bisection property:

1. Let chord $AB$ have midpoint $M$, and $OM \perp AB$
2. Draw radii $OA$ and $OB$ (both length $r$)
3. Triangles $OMA$ and $OMB$ are congruent by RHS (right angle at $M$, hypotenuse $OA = OB = r$, common side $OM$)
4. Therefore $AM = MB$ (corresponding parts of congruent triangles)
5. Why this step? Congruent triangles have all corresponding parts equal, so the chord halves must be equal.

Notation: Arc $AB$ is written as $\overset{\frown}{AB}$

Arc length formula: For a central angle $\theta$ (in radians): $\text{Arc length} = r\theta$

Derivation:

1. Full circle circumference: $C = 2\pi r$ (this comes from the definition of $\pi$ as the ratio of circumference to diameter)
2. A full circle corresponds to angle $2\pi$ radians
3. For angle $\theta$, the arc is $\frac{\theta}{2\pi}$ of the full circle
4. Arc length $= \frac{\theta}{2\pi} \times 2\pi r = r\theta$
5. Why this step? We're using proportional reasoning: the arc length is to the full circumference as the angle is to the full rotation.

Derivation from first principles:

1. Full circle area: $A = \pi r^2$
   • Why? Consider concentric circles approximating a filled disk. As we add infinitesimally thin rings of radius $r$ and thickness $dr$, each ring has circumference $2\pi r$ and area $2\pi r \, dr$. Integrating from $0$ to $r:$\int_0^r 2\pi r , dr = \pi r^2$
2. Full angle in circle: $2\pi$ radians
3. Sector with angle $\theta$ is fraction $\frac{\theta}{2\pi}$ of circle
4. Sector area $= \frac{\theta}{2\pi} \times \pi r^2 = \frac{1}{2}r^2\theta$
5. Why this step? Again, proportional reasoning: the sector area relates to full circle area as the angle relates to full rotation.

Special case: For $\theta$ in degrees, convert to radians first: $\theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}$, so: $A_{\text{sector}} = \frac{\theta_{\text{deg}}}{360} \times \pi r^2$

Derivation:

1. A segment is created when a chord cuts across sector
2. The sector area includes both the segment and a triangle with vertices at the centre and chord endpoints
3. To get just the segment, subtract the triangle: $A_{\text{segment}} = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta$
4. For angle $\theta$ at centre, the triangle has two sides of length $r$ (the radii) meeting at angle $\theta$
5. Triangle area formula for two sides $a$, $b$ with included angle $\theta$: $\frac{1}{2}ab\sin\theta$
6. Here: $A_{\text{triangle}} = \frac{1}{2}r \cdot r \cdot \sin\theta = \frac{1}{2}r^2\sin\theta$
7. Why this step? The triangle area formula comes from height $= r\sin\theta$ when one radius is the base.

Final formula: $A_{\text{segment}} = \frac{1}{2}r^2(\theta - \sin\theta)$

Worked Examples

Solution: (a) Diameter: $d = 2r = 2(5) = 10$ cm

• Why this step? Diameter is defined as twice the radius, passing through the centre.

(b) Circumference: $C = 2\pi r = 2\pi(5) = 10\pi \approx 31.42$ cm

• Why this step? Circumference formula comes from the definition of $\pi$.

(c) Area: $A = \pi r^2 = \pi(5)^2 = 25\pi \approx 78.54$ cm²

• Why this step? Area formula derived from integrating concentric rings.

Solution: First, convert angle to radians: $\theta = 60° \times \frac{\pi}{180} = \frac{\pi}{3}$ radians

• Why this step? Standard formulas require radians because $s = r\theta$ only works when $\theta$ is in radians.

(a) Arc length: $s = r\theta = 6 \times \frac{\pi}{3} = 2\pi \approx 6.28$ cm

• Why this step? Arc length is proportional to angle: $\frac{\theta}{2\pi}$ of full circumference.

(b) Sector area: $A = \frac{1}{2}r^2\theta = \frac{1}{2}(6)^2 \times \frac{\pi}{3} = 6\pi \approx 18.85$ cm²

• Why this step? Sector area is the same fraction $\frac{\theta}{2\pi}$ of the full circle area.

Verification: Sector should be $\frac{60°}{360°} = \frac{1}{6}$ of circle. Full area: $\pi(6)^2 = 36\pi$ cm² One-sixth: $\frac{36\pi}{6} = 6\pi$ cm² ✓

Solution: Angle in radians: $\theta = 90° \times \frac{\pi}{180} = \frac{\pi}{2}$ radians

• Why this step? Converting to radians for the formula.

Sector area: $A_{\text{sector}} = \frac{1}{2}r^2\theta = \frac{1}{2}(8)^2 \times \frac{\pi}{2} = 16\pi$ cm²

• Why this step? First find the full "pizza slice" area.

Triangle area: $A_{\text{triangle}} = \frac{1}{2}r^2\sin\theta = \frac{1}{2}(8)^2\sin(90°) = \frac{1}{2}(64)(1) = 32$ cm²

• Why this step? The triangle formed by two radii and the chord has area $\frac{1}{2}r^2\sin\theta$. For $90°$, $\sin(90°) = 1$.

Segment area: $A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = 16\pi - 32 \approx 50.27 - 32 = 18.27$ cm²

• Why this step? Remove the triangle part from the sector to get just the curved segment.

Solution: Let $M$ be the foot of perpendicular from centre $O$ to chord $AB$.

• $OM = 6$ cm (given)
• $OA = 10$ cm (radius)
• Triangle $OMA$ is right-angled at $M$

By Pythagorean theorem: $AM^2 = OA^2 - OM^2 = 10^2 - 6^2 = 100 - 36 = 64$ $AM = 8 \text{ cm}$

• Why this step? In the right triangle, the radius is hypotenuse, perpendicular distance and half-chord are the other two sides.

Since perpendicular from centre bisects chord: $AB = 2 \times AM = 2 \times 8 = 16 \text{ cm}$

• Why this step? The perpendicular bisection property: $OM \perp AB$ implies $AM = MB$.

Common Mistakes

Memory Aids

Visual memory: Imagine a pizza:

• Centre = where you hold it
• Radius = from centre to crust
• Diameter = full width across centre
• Chord = cutting across (not through centre)
• Arc = curved crust edge
• Sector = your slice (triangular)
• Segment = just the crust part of your slice

Now, if you draw any straight line connecting two points on the circle, that's a chord. The diameter is the biggest chord possible. The curved part of the circle between two points is an arc—like the crust of a pizza slice.

A sector is like a pizza slice with the pointy end at the centre—you have two radii (like knife cuts) and the curved crust. A segment is what's left if you cut off that pointy triangle part—just the curved crust section between a straight line (chord) and the circle's edge.

Why do we care? Circles are everywhere! Clock faces (radius to each hour mark), wheels (all spokes same length = radii), slicing cakes fairly (equal sector angles = equal areas). Understanding these parts lets you calculate exactly how much pizza you get, or how far you walk on a circular track, or how to design a round table!

Active Recall Practice

#flashcards/maths

What is the centre of a circle? :: The fixed point from which all points on the circle are equidistant. Symbol: $O$.

Why must angle be in radians for $s = r\theta$? :: Because radians are defined as arc length divided by radius ($\theta = \frac{s}{r}$), making the formula dimensionally consistent. Degrees are arbitrary units.

Connections

• Circle Equations — Using $(x-h)^2 + (y-k)^2 = r^2$ for circles in coordinate geometry
• Angle Properties in Circles — Inscribed angles, central angles, and their relationships to arcs
• Tangent and Secant Lines — Lines that touch or intersect circles
• Circle Theorems — Properties like "angle in semicircle is 90°"
• Circumference and Area of Circles — Detailed derivations of $C = 2\pi r$ and $A = \pi r^2$
• Pythagorean Theorem — Used extensively in chord-radius calculations
• Radian Measure — Why radians are natural units for circular measurements
• Sectors and Pizza Problem — Real-world applications of sector calculations
• Integration — How $A = \pi r^2$ is derived by integrating concentric rings

Study tip: Draw each part on an actual circle. Physical sketching with a compass helps internalize the relationships between radius, chord, and arc much better than just reading.

Concept Map

Hinglish (regional understanding)