Areas Related to Circles - Class 10 Maths

1. Circles, Radius & Centre

A circle is the path traced by a point that moves in a plane so that its distance from a fixed point always stays the same. That fixed point is called the centre, and the fixed distance is called the radius (denoted $r$).

Centre: the fixed point equidistant from every point on the circle.

Radius ($r$): the distance from the centre to any point on the circle. Every radius of the same circle has equal length.

2. Chord

A chord is a straight line segment joining any two points on the circle. A chord always lies entirely inside the circle (except its two endpoints, which sit on the circle itself).

3. Diameter

A diameter is a special chord that passes through the centre of the circle. It is the longest possible chord in any circle, and it's exactly twice the radius.

Formula $$d = 2r$$

4. Circumference

The circumference is the total length of the boundary of the circle — essentially, its perimeter. It's always a fixed multiple of the radius, given by the constant $\pi$ (pi ≈ 3.14159).

Formula $$\text{Circumference} = 2\pi r = \pi d$$

5. Secant

A secant is a line that intersects the circle at exactly two points, but — unlike a chord — it extends infinitely in both directions beyond the circle. A chord is really just the portion of a secant line that lies inside the circle. Tangent is also referred to as the limiting case of secant when both intersection point overlap each other

6. Arc

An arc is any part of the circumference of the circle — a curved "slice" of the boundary between two points on the circle. Any two points on a circle divide it into two arcs:

Minor arc: the shorter of the two arcs.

Major arc: the longer of the two arcs.

7. Central Angle

The central angle is the angle formed at the centre of the circle by two radii drawn to the endpoints of an arc (or a sector). It's what determines how "big" a sector or arc is — a full circle corresponds to a central angle of $360°$.

Minor arc: central angle less than 180˚.

Major arc: central angle more than 180˚.

8. Semicircle

A semicircle is exactly half of a circle, formed when a diameter divides the circle into two equal parts. The arc of a semicircle always corresponds to a central angle of $180°$.

9. Segment

A segment is the region enclosed between a chord and the arc it cuts off — it's what's "left over" between a straight chord and the curved boundary. Every chord creates two segments (diameter creates two semicircle):

Minor segment: the smaller region, cut off by the minor arc.

Major segment: the larger region, cut off by the major arc.

10. Sector

A sector is the region enclosed by two radii and the arc between them — think of a single "slice" of a circular pizza. Like an arc, every pair of radii creates a minor sector (smaller slice) and a major sector (larger slice), unless the two radii are exactly opposite, giving two semicircular sectors.

Minor sector: the smaller region, cut off by the minor arc.

Major sector: the larger region, cut off by the major arc.

11. Quadrant

A quadrant is simply a special sector whose central angle is exactly $90°$ — in other words, one quarter of the full circle. Four quadrants make up a complete circle.

12. Circumcircle & Circumradius

The circumcircle of a polygon (most commonly a triangle) is a circle that passes through every vertex of that polygon — the polygon is said to be "inscribed" in the circle. The radius of this circumcircle is called the circumradius.

13. Incircle & Inradius

The incircle of a polygon is the largest circle that fits entirely inside it, touching every one of its sides exactly once. The radius of this incircle is called the inradius. Each of the sides of polygon is a tangent to the circle

14. Key Formulas

All of the following use $r$ for radius and $\theta$ for the central angle in degrees.

Area of a Circle $$A = \pi r^2$$

Circumference of a Circle $$C = 2\pi r$$

Area of a Sector $$A_{sector} = \frac{\theta}{360°} \times \pi r^2$$

Length of an Arc $$\text{Arc length} = \frac{\theta}{360°} \times 2\pi r$$

Area of a Segment (minor segment) $$A_{segment} = A_{sector} - A_{triangle} = \frac{\theta}{360°} \times \pi r^2 - \frac{1}{2}r^2\sin\theta$$

Area & Perimeter of a Semicircle $$A_{semicircle} = \frac{1}{2}\pi r^2 \qquad \text{Perimeter} = \pi r + 2r$$

15. Rotation of Clock Hands

Since a clock face is a circle divided into 12 hours and 60 minutes, the hands' movement is really a sector-rotation problem in disguise.

Minute hand: completes a full $360°$ rotation in 60 minutes → moves $6°$ per minute.

Hour hand: completes a full $360°$ rotation in 12 hours (720 minutes) → moves $0.5°$ per minute.

Area swept by the minute hand in $t$ minutes $$A = \frac{6t}{360°} \times \pi r^2 = \frac{t}{60} \times \pi r^2$$

16. Rotation of a Wheel

In one complete rotation, a wheel covers a distance exactly equal to its own circumference. This connects circle geometry directly to speed-and-distance problems.

Distance in one rotation $$\text{Distance} = 2\pi r$$

Number of rotations for a given distance $$\text{Rotations} = \frac{\text{Total distance}}{2\pi r}$$

17. Touching Circles

When two circles touch each other at exactly one point, the relationship between their radii and the distance between their centres follows a simple rule depending on how they touch:

Touching externally (circles outside each other, touching at one point): distance between centres $= r_1 + r_2$.

Touching internally (one circle inside the other, touching at one point): distance between centres $= |r_1 - r_2|$.

18. Solved Examples

Carefully selected standard problems covering board-level examination patterns with comprehensive step-by-step working and accompanying visual diagrams.

Example 1 • Area to Circumference

The area of a circular play ring is $616\text{ cm}^2$. Find its circumference. (Use $\pi = \frac{22}{7}$)

Solution:
Let the radius of the circle be $r\text{ cm}$.
Given: $$\text{Area} = \pi r^2 = 616$$ $$\frac{22}{7} \times r^2 = 616 \implies r^2 = \frac{616 \times 7}{22} = 28 \times 7 = 196$$ $$r = \sqrt{196} = 14\text{ cm}$$ Now, circumference of the circle: $$\text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88\text{ cm}$$ Thus, the circumference of the play ring is $88\text{ cm}$.

Example 2 • Semicircle Perimeter to Area

The perimeter of a semicircular protractor is $36\text{ cm}$. Calculate its total area. (Use $\pi = \frac{22}{7}$)

Solution:
The perimeter of a closed semicircle consists of its curved arc plus its straight diameter: $$\text{Perimeter} = \pi r + 2r = r(\pi + 2)$$ $$r\left(\frac{22}{7} + 2\right) = 36 \implies r\left(\frac{36}{7}\right) = 36 \implies r = 7\text{ cm}$$ Now, find its area: $$\text{Area} = \frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77\text{ cm}^2$$ Thus, the area of the protractor is $77\text{ cm}^2$.

Example 3 • Sector Angle to Chord Length

A sector of a circle of radius $10.5\text{ cm}$ subtends an angle of $60^\circ$ at the centre $O$. Find the length of the corresponding chord $AB$.

Solution:
In $\triangle OAB$, $OA = OB = 10.5\text{ cm}$ (radii of the same circle).
Since $OA = OB$, the angles opposite to these sides are equal: $$\angle OAB = \angle OBA$$ Given central angle $\angle AOB = 60^\circ$: $$\angle OAB + \angle OBA = 180^\circ - 60^\circ = 120^\circ \implies \angle OAB = \angle OBA = 60^\circ$$ Since all three internal angles are $60^\circ$, $\triangle OAB$ is an equilateral triangle.
Therefore: $$\text{Chord } AB = OA = OB = 10.5\text{ cm}$$

Example 4 • Ratio of Radii to Perimeter

The radii of two semicircular regions are in the ratio $3 : 5$. Find the ratio of their perimeters.

Solution:
Let the radii of the two semicircles be $r_1$ and $r_2$, such that $\frac{r_1}{r_2} = \frac{3}{5}$.
The perimeter of a closed semicircle of radius $r$ is: $$P = \pi r + 2r = r(\pi + 2)$$ Taking the ratio of their perimeters: $$\frac{P_1}{P_2} = \frac{r_1(\pi + 2)}{r_2(\pi + 2)} = \frac{r_1}{r_2} = \frac{3}{5}$$ Thus, the ratio of their perimeters is $3 : 5$.

Example 5 • Wheel Revolutions

A car travels at a uniform speed of $66\text{ km/h}$. If the diameter of each wheel is $70\text{ cm}$, find the number of complete revolutions each wheel makes in $10\text{ minutes}$. (Use $\pi = \frac{22}{7}$)

Solution:
1. Radius of wheel $r = \frac{70}{2} = 35\text{ cm} = 0.35\text{ m}$.
2. Distance covered in $1$ revolution (Circumference): $$C = 2\pi r = 2 \times \frac{22}{7} \times 35 = 220\text{ cm} = 2.2\text{ m}$$ 3. Speed in metres per minute: $$\text{Speed} = \frac{66 \times 1000\text{ m}}{60\text{ min}} = 1100\text{ m/min}$$ 4. Total distance covered in $10\text{ minutes}$: $$D = 1100 \times 10 = 11,000\text{ m}$$ 5. Number of complete revolutions: $$\text{Revolutions} = \frac{\text{Total Distance}}{\text{Circumference}} = \frac{11,000}{2.2} = 5,000\text{ revolutions}$$

Example 6 • Hour Hand Movement

The length of the hour hand of a wall clock is $4.2\text{ cm}$. Find the area swept by it between $7:00\text{ AM}$ and $9:20\text{ AM}$. (Use $\pi = \frac{22}{7}$)

Solution:
1. Time duration from $7:00\text{ AM}$ to $9:20\text{ AM} = 2\text{ h } 20\text{ min} = 2 + \frac{20}{60} = \frac{7}{3}\text{ hours}$.
2. In $12\text{ hours}$, the hour hand describes an angle of $360^\circ$: $$\text{Angle described per hour} = \frac{360^\circ}{12} = 30^\circ$$ $$\text{Angle swept } \theta = \frac{7}{3} \times 30^\circ = 70^\circ$$ 3. Area swept (Area of sector with $r = 4.2\text{ cm}$): $$\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{70}{360} \times \frac{22}{7} \times (4.2)^2 = 10.78\text{ cm}^2$$

Example 7 • Wire: Square to Sector

A wire is bent into the shape of a square of side $11\text{ cm}$. The same wire is unbent and reshaped into a circular sector of radius $7\text{ cm}$. Find the central angle subtended by the sector. (Use $\pi = \frac{22}{7}$)

Solution:
1. Total length of wire = Perimeter of square: $$L = 4 \times 11 = 44\text{ cm}$$ 2. Perimeter of a sector consists of 2 radii and arc length $l$: $$\text{Perimeter} = 2r + l \implies 44 = 2(7) + l \implies l = 44 - 14 = 30\text{ cm}$$ 3. Arc length formula: $$l = \frac{\theta}{360^\circ} \times 2\pi r \implies 30 = \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 7$$ $$30 = \frac{\theta}{360^\circ} \times 44 \implies \theta = \frac{30 \times 360}{44} = \frac{2700}{11} \approx 245.45^\circ\text{ (or } 245\frac{5}{11}^\circ)$$

Example 8 • Outer Circular Path & Levelling

A circular flower bed of radius $28\text{ m}$ is surrounded by an outer concentric path of uniform width $3.5\text{ m}$. Find the total cost of levelling the path at ₹$25\text{ per m}^2$. (Use $\pi = \frac{22}{7}$)

Solution:
Inner radius $r = 28\text{ m}$
Outer radius $R = 28 + 3.5 = 31.5\text{ m}$
Area of the circular path: $$\text{Area} = \pi(R^2 - r^2) = \pi(R - r)(R + r)$$ $$\text{Area} = \frac{22}{7} \times (31.5 - 28) \times (31.5 + 28) = \frac{22}{7} \times 3.5 \times 59.5 = 11 \times 59.5 = 654.5\text{ m}^2$$ Total cost at ₹$25\text{ per m}^2$: $$\text{Cost} = 654.5 \times 25 = \text{₹}16,362.50$$

Example 9 • Inner Circular Path & Gravelling

A circular lawn has a diameter of $42\text{ m}$. A walking track of width $3.5\text{ m}$ is constructed along the inside boundary of the lawn. Find the cost of gravelling the track at ₹$40\text{ per m}^2$. (Use $\pi = \frac{22}{7}$)

Solution:
Outer radius $R = \frac{42}{2} = 21\text{ m}$
Inner radius $r = 21 - 3.5 = 17.5\text{ m}$
Area of the inner track: $$\text{Area} = \pi(R^2 - r^2) = \frac{22}{7} \times (21 - 17.5) \times (21 + 17.5)$$ $$\text{Area} = \frac{22}{7} \times 3.5 \times 38.5 = 11 \times 38.5 = 423.5\text{ m}^2$$ Total cost at ₹$40\text{ per m}^2$: $$\text{Cost} = 423.5 \times 40 = \text{₹}16,940$$

Example 10 • Percentage Increase in Area

If the radius of a circle is increased by $20\%$, find the percentage increase in its area.

Solution:
Let the original radius be $r$. Original area $A_1 = \pi r^2$.
New radius after $20\%$ increase: $$r' = r + \frac{20}{100}r = 1.2r$$ New area: $$A_2 = \pi (r')^2 = \pi (1.2r)^2 = 1.44 \pi r^2$$ Increase in area: $$\Delta A = A_2 - A_1 = 1.44\pi r^2 - \pi r^2 = 0.44\pi r^2$$ Percentage increase: $$\text{Percentage Increase} = \left(\frac{0.44\pi r^2}{\pi r^2}\right) \times 100\% = 44\%$$

Example 11 • Percentage Decrease in Area

If the circumference of a circle is decreased by $30\%$, by what percentage does its area decrease?

Solution:
Since Circumference $C = 2\pi r$, circumference varies directly with radius ($C \propto r$). Hence, a $30\%$ reduction in circumference indicates a $30\%$ reduction in radius.
Let initial radius be $r$. Then new radius: $$r' = r - 0.30r = 0.7r$$ Initial Area $A_1 = \pi r^2$, and new area: $$A_2 = \pi (0.7r)^2 = 0.49\pi r^2$$ Decrease in area: $$\Delta A = A_1 - A_2 = \pi r^2 - 0.49\pi r^2 = 0.51\pi r^2$$ Percentage decrease: $$\text{Percentage Decrease} = \left(\frac{0.51\pi r^2}{\pi r^2}\right) \times 100\% = 51\%$$

Example 12 • Complex Figure: Square with 4 Corner Quadrants

From each corner of a square of side $14\text{ cm}$, a quadrant of a circle of radius $3.5\text{ cm}$ is cut away, and a circle of diameter $7\text{ cm}$ is also removed from the centre. Find the area of the remaining shaded region. (Use $\pi = \frac{22}{7}$)

Solution:
1. Area of square of side $a = 14\text{ cm}$: $$\text{Area of square} = a^2 = 14 \times 14 = 196\text{ cm}^2$$ 2. Four quadrants of radius $r = 3.5\text{ cm}$ together form $1$ full circle: $$\text{Area of } 4\text{ quadrants} = 4 \times \left(\frac{1}{4}\pi r^2\right) = \frac{22}{7} \times 3.5 \times 3.5 = 38.5\text{ cm}^2$$ 3. Area of central circle of diameter $7\text{ cm}$ ($r_c = 3.5\text{ cm}$): $$\text{Area of central circle} = \pi r_c^2 = \frac{22}{7} \times 3.5 \times 3.5 = 38.5\text{ cm}^2$$ 4. Remaining shaded area: $$\text{Remaining Area} = \text{Area of square} - (\text{Area of } 4\text{ quadrants} + \text{Area of central circle})$$ $$\text{Remaining Area} = 196 - (38.5 + 38.5) = 196 - 77 = 119\text{ cm}^2$$

Example 13 • Complex Figure: Overlapping Semicircles (Leaf Pattern)

In a square $ABCD$ of side $10\text{ cm}$, semicircles are drawn on sides $AB$ and $AD$ as diameters. Find the area of the shaded leaf/petal region common to both semicircles. (Use $\pi = 3.14$)

Solution:
Let side of square $a = 10\text{ cm}$.
Diameter of each semicircle $= 10\text{ cm} \implies \text{radius } r = 5\text{ cm}$.
The area of the overlapping petal formed by two semicircles on adjacent sides is: $$\text{Area} = r^2\left(\frac{\pi}{2} - 1\right) = 5^2\left(\frac{3.14}{2} - 1\right)$$ $$\text{Area} = 25 \times (1.57 - 1) = 25 \times 0.57 = 14.25\text{ cm}^2$$

Example 14 • Complex Figure: Area between 3 Touching Circles

Three identical coins of radius $7\text{ cm}$ each are placed on a flat table such that each circle touches the other two externally. Calculate the area of the region enclosed between them. (Use $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.732$)

Solution:
1. Joining the centres of the three circles creates an equilateral triangle with side length: $$a = r + r = 7 + 7 = 14\text{ cm}$$ 2. Area of this equilateral triangle: $$\text{Area}_{\triangle} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 14 \times 14 = 49\sqrt{3} = 49 \times 1.732 = 84.868\text{ cm}^2$$ 3. The three sectors inside the triangle each have a central angle $\theta = 60^\circ$: $$\text{Combined Sector Area} = 3 \times \left(\frac{60^\circ}{360^\circ} \times \pi r^2\right) = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77\text{ cm}^2$$ 4. Area enclosed between the circles: $$\text{Area} = \text{Area}_{\triangle} - \text{Area of 3 sectors}$$ $$\text{Area} = 84.868 - 77 = 7.868\text{ cm}^2 \approx 7.87\text{ cm}^2$$

Example 15 • Circles Touching Externally

Two circles touch each other externally. The sum of their areas is $130\pi\text{ cm}^2$ and the distance between their centres is $14\text{ cm}$. Find the radii of the two circles.

Solution:
Let the radii of the two circles be $r_1$ and $r_2$.
1. When two circles touch externally, the distance between their centres equals the sum of their radii: $$r_1 + r_2 = 14 \implies r_2 = 14 - r_1$$ 2. Sum of their areas is given as $130\pi\text{ cm}^2$: $$\pi r_1^2 + \pi r_2^2 = 130\pi \implies r_1^2 + r_2^2 = 130$$ 3. Substitute $r_2 = 14 - r_1$: $$r_1^2 + (14 - r_1)^2 = 130$$ $$r_1^2 + 196 - 28r_1 + r_1^2 = 130$$ $$2r_1^2 - 28r_1 + 66 = 0 \implies r_1^2 - 14r_1 + 33 = 0$$ 4. Factoring the quadratic: $$(r_1 - 11)(r_1 - 3) = 0 \implies r_1 = 11\text{ or } r_1 = 3$$ If $r_1 = 11\text{ cm}$, then $r_2 = 14 - 11 = 3\text{ cm}$.
Thus, the radii of the two circles are $11\text{ cm}$ and $3\text{ cm}$.