Complete Guide to Area of a Circle
Master everything about circles! This comprehensive guide covers all essential formulas, theorems, and calculations for circles including area, circumference, sectors, segments, and semicircles. Perfect for students in Class 10, GCSE, IGCSE, IB, and AP curricula. Learn through detailed explanations, derivations, interactive calculators, and worked examples.
Circle Calculator
What is a Circle?
A circle is a closed two-dimensional shape where all points on the boundary are equidistant from a fixed central point called the center. This equal distance from the center to any point on the circle is called the radius. Circles are fundamental shapes in mathematics and appear throughout nature, architecture, and everyday life.
Key Properties of a Circle:
- A circle is a perfectly symmetrical shape
- It has infinite lines of symmetry passing through the center
- A circle has no corners or edges
- The circle is not a polygon (it has a curved boundary, not straight sides)
- The value of π (pi) ≈ 3.14159 is fundamental to all circle calculations
Important Terms Related to Circles
Radius (r)
Distance from center to any point on the circle
Diameter (d)
Distance across the circle through the center
\( d = 2r \)
Circumference (C)
The perimeter or distance around the circle
Chord
A line segment connecting two points on the circle
Arc
A portion of the circumference
Sector
A "pizza slice" region bounded by two radii and an arc
Area of a Circle Formula
The area of a circle is the amount of space enclosed within its boundary. It is measured in square units.
Primary Formula (Using Radius):
\[ A = \pi r^2 \]
Where:
- \( A \) = Area of the circle (square units)
- \( r \) = Radius of the circle (linear units)
- \( \pi \) = Pi (approximately 3.14159 or \( \frac{22}{7} \))
Alternative Formulas:
Using Diameter:
\[ A = \frac{\pi d^2}{4} \]
Using Circumference:
\[ A = \frac{C^2}{4\pi} \]
Derivation of Circle Area Formula
The formula \( A = \pi r^2 \) can be derived using several methods. Here are two classical approaches:
Method 1: Using Concentric Circles and Triangles
Step-by-step derivation:
- Imagine dividing the circle into many thin concentric rings
- Cut these rings and straighten them out to form triangular shapes
- When arranged, they approximate a triangle with:
- Base = Circumference = \( 2\pi r \)
- Height = Radius = \( r \)
- Area of triangle: \( A = \frac{1}{2} \times \text{base} \times \text{height} \)
- Substituting: \( A = \frac{1}{2} \times 2\pi r \times r = \pi r^2 \)
Method 2: Using Regular Polygons
As the number of sides of a regular polygon inscribed in a circle increases, the polygon approaches the shape of the circle:
- For a regular polygon with \( n \) sides, each forming an isosceles triangle with the center
- Area of each triangle = \( \frac{1}{2} \times s \times h \) where \( s \) is the side length and \( h \) is the height
- Total area = \( n \times \frac{1}{2} \times s \times h = \frac{1}{2}(ns)h \)
- As \( n \to \infty \): \( ns \to 2\pi r \) (circumference) and \( h \to r \)
- Therefore: \( A = \frac{1}{2} \times 2\pi r \times r = \pi r^2 \)
Circumference of a Circle
The circumference is the perimeter or distance around the circle. It is the one-dimensional measure of the circle's boundary.
Circumference Formulas:
Using Radius:
\[ C = 2\pi r \]
Using Diameter:
\[ C = \pi d \]
Relationship with Area:
\[ r = \sqrt{\frac{A}{\pi}} \quad \text{then} \quad C = 2\pi\sqrt{\frac{A}{\pi}} \]
Diameter and Radius Relationships
Diameter from Radius:
\[ d = 2r \]
Radius from Diameter:
\[ r = \frac{d}{2} \]
Radius from Area:
\[ r = \sqrt{\frac{A}{\pi}} \]
Radius from Circumference:
\[ r = \frac{C}{2\pi} \]
Area of a Sector
A sector is a region of a circle bounded by two radii and the arc between them. It resembles a "slice of pizza" or "piece of pie."
Sector Area Formula (Central Angle in Degrees):
\[ A_{\text{sector}} = \frac{\theta}{360°} \times \pi r^2 \]
Sector Area Formula (Central Angle in Radians):
\[ A_{\text{sector}} = \frac{1}{2}r^2\theta \]
Where:
- \( \theta \) = Central angle (the angle at the center of the circle)
- \( r \) = Radius of the circle
Derivation:
The sector area is proportional to the angle it subtends at the center:
\[ \frac{\text{Sector Area}}{\text{Circle Area}} = \frac{\text{Sector Angle}}{\text{Full Angle}} \]
\[ \frac{A_{\text{sector}}}{\pi r^2} = \frac{\theta}{360°} \]
Arc Length of a Sector
Arc Length (Degrees):
\[ L = \frac{\theta}{360°} \times 2\pi r \]
Arc Length (Radians):
\[ L = r\theta \]
Segment of a Circle
A segment is the region bounded by a chord and the arc subtended by the chord. There are two types: minor segment (smaller area) and major segment (larger area).
Area of Segment Formula:
\[ A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} \]
\[ A_{\text{segment}} = \frac{1}{2}r^2(\theta - \sin\theta) \]
Where \( \theta \) is in radians
For degrees:
\[ A_{\text{segment}} = \frac{\pi r^2\theta}{360} - \frac{1}{2}r^2\sin\theta \]
Area of a Semicircle
A semicircle is exactly half of a circle, formed by cutting a circle along its diameter.
Semicircle Area Formula:
\[ A_{\text{semicircle}} = \frac{\pi r^2}{2} \]
Using Diameter:
\[ A_{\text{semicircle}} = \frac{\pi d^2}{8} \]
Perimeter of Semicircle:
The perimeter includes the curved part (half circumference) plus the diameter:
\[ P_{\text{semicircle}} = \pi r + 2r = r(\pi + 2) \]
Circle Theorems
Circle theorems describe important relationships between angles, chords, tangents, and other parts of circles. These are essential for geometry problems.
Key Circle Theorems
1. Angle at the Centre Theorem:
The angle subtended by an arc at the center is twice the angle subtended at any point on the circumference.
\[ \text{Central Angle} = 2 \times \text{Inscribed Angle} \]
2. Angle in a Semicircle Theorem:
Any angle inscribed in a semicircle (where the diameter is one side) is always a right angle (90°).
\[ \text{Angle in semicircle} = 90° \]
3. Angles in the Same Segment:
Angles inscribed in the same segment of a circle (subtended by the same arc) are equal.
4. Cyclic Quadrilateral Theorem:
In a cyclic quadrilateral (all vertices on a circle), opposite angles sum to 180°.
\[ \angle A + \angle C = 180° \quad \text{and} \quad \angle B + \angle D = 180° \]
5. Tangent-Radius Theorem:
A tangent to a circle is perpendicular to the radius at the point of contact.
6. Alternate Segment Theorem:
The angle between a tangent and a chord equals the angle in the alternate segment.
Central Angle in a Circle
A central angle is an angle whose vertex is at the center of the circle and whose sides are radii extending to the circumference.
Total Degrees in a Circle:
\[ \text{Full circle} = 360° = 2\pi \text{ radians} \]
Converting Between Degrees and Radians:
\[ \text{Radians} = \frac{\pi}{180°} \times \text{Degrees} \]
\[ \text{Degrees} = \frac{180°}{\pi} \times \text{Radians} \]
Common Circle Size Examples
Here are calculations for circles with common inch measurements:
| Diameter | Radius | Area (exact) | Area (approx) | Circumference |
|---|---|---|---|---|
| 6 inches | 3 in | \( 9\pi \) in² | 28.27 in² | 18.85 in |
| 8 inches | 4 in | \( 16\pi \) in² | 50.27 in² | 25.13 in |
| 9 inches | 4.5 in | \( 20.25\pi \) in² | 63.62 in² | 28.27 in |
| 10 inches | 5 in | \( 25\pi \) in² | 78.54 in² | 31.42 in |
| 12 inches | 6 in | \( 36\pi \) in² | 113.10 in² | 37.70 in |
| 14 inches | 7 in | \( 49\pi \) in² | 153.94 in² | 43.98 in |
Worked Examples
Example 1: Finding Area from Radius
Problem: Find the area of a circle with radius 7 cm.
Solution:
Using the formula \( A = \pi r^2 \)
\[ A = \pi \times 7^2 = 49\pi \text{ cm}^2 \]
\[ A \approx 153.94 \text{ cm}^2 \]
Example 2: Finding Area from Diameter
Problem: A circle has a diameter of 12 inches. Find its area.
Solution:
First find the radius: \( r = \frac{d}{2} = \frac{12}{2} = 6 \) inches
\[ A = \pi r^2 = \pi \times 6^2 = 36\pi \text{ in}^2 \]
\[ A \approx 113.10 \text{ in}^2 \]
Example 3: Finding Sector Area
Problem: Find the area of a sector with central angle 60° and radius 10 cm.
Solution:
Using \( A_{\text{sector}} = \frac{\theta}{360°} \times \pi r^2 \)
\[ A_{\text{sector}} = \frac{60°}{360°} \times \pi \times 10^2 = \frac{1}{6} \times 100\pi \]
\[ A_{\text{sector}} = \frac{100\pi}{6} = \frac{50\pi}{3} \approx 52.36 \text{ cm}^2 \]
Example 4: Finding Radius from Circumference
Problem: A circle has a circumference of 31.4 cm. Find its radius and area.
Solution:
Using \( C = 2\pi r \), we get \( r = \frac{C}{2\pi} \)
\[ r = \frac{31.4}{2\pi} = \frac{31.4}{6.28} \approx 5 \text{ cm} \]
Now find the area:
\[ A = \pi r^2 = \pi \times 5^2 = 25\pi \approx 78.54 \text{ cm}^2 \]
Example 5: Semicircle Area
Problem: Find the area of a semicircle with diameter 14 m.
Solution:
Radius: \( r = \frac{14}{2} = 7 \) m
\[ A_{\text{semicircle}} = \frac{\pi r^2}{2} = \frac{\pi \times 7^2}{2} = \frac{49\pi}{2} \]
\[ A_{\text{semicircle}} \approx 76.97 \text{ m}^2 \]
Special Topics and Applications
Squaring the Circle
Squaring the circle is a famous impossible problem from ancient geometry: construct a square with the same area as a given circle using only compass and straightedge.
This was proven impossible in 1882 because π is transcendental (cannot be the root of any polynomial equation with rational coefficients).
Relationship: If a circle has radius \( r \), a square with the same area would have side length:
\[ s = r\sqrt{\pi} \]
Circles and Triangles
Circumcircle: A circle that passes through all vertices of a triangle
Incircle: A circle inscribed within a triangle, touching all three sides
Circumradius Formula:
\[ R = \frac{abc}{4K} \]
Where \( a, b, c \) are the side lengths and \( K \) is the area of the triangle.
Inradius Formula:
\[ r = \frac{K}{s} \]
Where \( K \) is the area and \( s \) is the semi-perimeter.
Equation of a Circle
Standard Form (center at origin):
\[ x^2 + y^2 = r^2 \]
General Form (center at (h, k)):
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Expanded General Form:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Where center is \( (-g, -f) \) and radius is \( \sqrt{g^2 + f^2 - c} \)
Practice Problems
Problem 1: A circular table cover has a radius of 1.5 m. Find its area and the length of lace required to border it.
Problem 2: Find the area of a sector of angle 45° in a circle of radius 8 cm.
Problem 3: The circumference of a circle is 44 cm. Find its area.
Problem 4: A circular park has a diameter of 140 m. Find the cost of fencing it at ₹20 per meter.
Problem 5: Find the perimeter of a semicircle with radius 10 cm.
Summary of Key Formulas
| Measurement | Formula | Notes |
|---|---|---|
| Area of Circle | \( A = \pi r^2 \) | Fundamental formula |
| Circumference | \( C = 2\pi r = \pi d \) | Perimeter of circle |
| Sector Area | \( A = \frac{\theta}{360°} \pi r^2 \) | θ in degrees |
| Semicircle Area | \( A = \frac{\pi r^2}{2} \) | Half of circle |
| Arc Length | \( L = \frac{\theta}{360°} \times 2\pi r \) | θ in degrees |
| Segment Area | \( A = \frac{1}{2}r^2(\theta - \sin\theta) \) | θ in radians |
| Semicircle Perimeter | \( P = r(\pi + 2) \) | Curved + straight parts |