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ICSE • Class 8 • Mathematics • Ch 21
Estimated Time: 45 Mins
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Circle

In ICSE Class 8 Mathematics, "Circle" provides an authoritative, geometrically rigorous master study guide investigating the geometry of circles, circular components, angle properties, chord theorems, and circle construction. This comprehensive chapter explores Fundamental Terminology (Center $O$, Radius $r$, Diameter $d = 2r$, Circumference $C = 2\pi r$; Interior, Boundary, and Exterior regions), Circular Lines & Segments (Secant: a straight line intersecting a circle at two distinct points; Tangent: a straight line touching a circle at exactly one point called the point of contact; Chord: line segment joining any two points on circumference; Diameter as the longest chord; Concentric Circles: circles sharing the identical center but having different radii; Congruent Circles: circles having equal radii), Circular Arcs & Sectors (Arc: continuous portion of circumference; Minor Arc vs Major Arc; Semicircle: half-circumference subtended by a diameter; Central Angle $\theta$; Sector: region bounded by two radii and the intercepted arc; Minor sector vs Major sector; Segment: region bounded by a chord and its intercepted arc; Minor segment vs Major segment), Fundamental Theorems on Chords of a Circle: 1. The perpendicular drawn from the center of a circle to a chord bisects the chord ($OM \perp AB \implies AM = MB = \frac{AB}{2}$), 2. The line joining the center to the midpoint of a chord is perpendicular to the chord, 3. Equal chords are equidistant from the center ($AB = CD \implies OM = ON$), 4. Chords equidistant from the center are equal in length, 5. Semicircle Theorem (The angle subtended by a diameter in a semicircle is strictly a right angle: $\angle ACB = 90^{\circ}$), and Circle Constructions using compass and ruler aligned with the 2026–27 CISCE ICSE curriculum.

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Why Is Every Manhole Cover in the World Shaped Like a Circle Instead of a Square?

Walk down any city street in London, Tokyo, or New York, and look at the heavy cast-iron manhole covers on the street. Every single one is a PERFECT CIRCLE! Have you ever wondered why? If a manhole cover were a square, a maintenance worker could tilt it diagonally and accidentally drop it right down the hole—because the diagonal of a square is $\sqrt{2} \approx 1.414$ times wider than its side! But a CIRCLE has a constant diameter ($d = 2r$) in EVERY SINGLE DIRECTION! It is mathematically impossible for a circular manhole cover to fall through its own circular hole, no matter how it is turned! Circles are the most symmetric and harmonious shapes in all of mathematics: every point on the circumference sits at the exact same distance $r$ from the center. What happens if you drop a perpendicular line from the center onto any chord? Why is an angle inscribed in a semicircle always exactly $90^{\circ}$? Let's master the circle.

Why This Chapter Matters

Circles are fundamental to engineering mechanics (wheels, gears, ball bearings, flywheels), celestial orbital dynamics (planetary motion, satellite orbits), telecommunications radar sweeps, and architecture (domes, arches). Mastering circle theorems and chord bisections is essential for ICSE Class 8, 9, and 10 Euclidean geometry.

Before You Begin (Prerequisites)

  • Basic geometric terms: point, line segment, angle.
  • Pythagoras theorem ($a^2 + b^2 = c^2$).
  • Circumference of circle ($C = 2\pi r$).

What You Will Learn (Core Objectives)

  • Define circle, radius, diameter, chord, secant, tangent, and concentric circles.
  • Distinguish between minor/major arcs, sectors, and segments.
  • Apply the Chord Bisection Theorem ($OM \perp AB \implies AM = MB$) with Pythagoras theorem.
  • Apply the theorem: Equal chords are equidistant from the center.
  • Utilize the Semicircle Right-Angle Theorem ($\angle ACB = 90^{\circ}$).
  • Calculate chord lengths, perpendicular distances, and radii using right-angled triangle relations.

Chapter Roadmap & Progression

1 1. Anatomy of a Circle: Lines, Regi...
2 2. Fundamental Chord Theorems
3 3. The Right-Angled Triangle Metric...
4 4. The Semicircle Right-Angle Theor...

Complete Concept Guide (100% Curriculum Coverage)

1. Anatomy of a Circle: Lines, Regions & Sectors

Understand
A. Fundamental Definitions:
  • Circle: The locus of all points in a plane equidistant from a fixed point called the Center ($O$). The fixed distance is the Radius ($r$).
  • Diameter ($d = 2r$): A chord that passes through the center. It is the longest possible chord of a circle.
  • Secant: A straight line that intersects a circle at two distinct points, cutting through the boundary.
  • Tangent: A straight line that touches the circle at only one single point (the point of contact).
B. Sectors vs Segments:
  • Sector: The interior region enclosed by two radii and the intercepted arc (like a slice of pizza!).
  • Segment: The interior region enclosed by a chord and the intercepted arc (like a bow and arrow!).

2. Fundamental Chord Theorems

Chord Theorems
Theorem 1: The Perpendicular Bisector of a Chord:

The perpendicular drawn from the center of a circle to a chord bisects the chord:

$$\mathbf{OM \perp AB \implies AM = MB = \frac{1}{2}AB}$$

Converse: The straight line joining the center of a circle to the midpoint of a chord is perpendicular to the chord.

Theorem 2: Chords & Distance from Center:
  1. Equal chords of a circle are equidistant from the center ($AB = CD \implies OM = ON$).
  2. Chords that are equidistant from the center of a circle are equal in length.
  3. Of any two unequal chords, the longer chord is closer to the center!

3. The Right-Angled Triangle Metric: $r^2 = d^2 + (c/2)^2$

Calculations
The Master Pythagorean Chord Identity:

Let a chord of length $c = AB$ lie at a perpendicular distance $d = OM$ from the center $O$, and let radius be $r = OA$. In right-angled triangle $\triangle OMA$:

$$\mathbf{r^2 = d^2 + \left( \frac{c}{2} \right)^2} \quad \Longleftrightarrow \quad \mathbf{c = 2\sqrt{r^2 - d^2}}$$

Example: If radius $r = 10\text{ cm}$ and chord length is $16\text{ cm}$, find distance from center:
Half-chord $= \frac{16}{2} = 8\text{ cm}$.
$$d^2 = r^2 - \left(\frac{c}{2}\right)^2 = 10^2 - 8^2 = 100 - 64 = 36 \implies \mathbf{d = 6\text{ cm}}$$.

4. The Semicircle Right-Angle Theorem

Angle in Semicircle
A. Inscribed Angle Theorem:

An angle inscribed in a semicircle is ALWAYS a right angle ($90^{\circ}$):

$$\mathbf{\text{If } AB \text{ is a diameter and } C \text{ is any point on the arc, then } \angle ACB = 90^{\circ}}$$

Geometric Proof: Join $O$ to $C$. Since $OA = OB = OC = r$, triangles $\triangle OAC$ and $\triangle OBC$ are isosceles. Let $\angle OAC = \angle OCA = x$ and $\angle OBC = \angle OCB = y$. In $\triangle ABC$: $x + y + (x + y) = 180^{\circ} \implies 2(x + y) = 180^{\circ} \implies x + y = \mathbf{90^{\circ}}$!

Key Formulas, Identities & Theorems

Pythagorean Chord-Radius Law
$$r^2 = OM^2 + AM^2 = d^2 + \left(\frac{c}{2}\right)^2$$
Connects radius r, chord length c, and distance d.
Angle in a Semicircle
$$\angle ACB = 90^{\circ} \quad [AB \text{ is diameter}]$$
Any triangle inscribed in a semicircle with diameter as hypotenuse is a right triangle.

Geometry: Chord Perpendicular Bisector & Semicircle Theorem

Circle Geometry: Chord Bisector Theorem & Semicircle Right Angle CHORD PERPENDICULAR BISECTOR O A B d M r OM ⊥ AB ⇒ AM = MB = AB / 2 r2 = d2 + (AB/2)2 (Pythagoras Theorem!) ANGLE IN A SEMICIRCLE IS 90° A B O C ∠ACB = 90° (Right Angle!) Hypotenuse is always the diameter AB PERPENDICULAR FROM CENTER BISECTS CHORD • r^2 = d^2 + (c/2)^2 • ∠ IN SEMICIRCLE = 90°

Chapter Summary & 10 Key Takeaways

Takeaway 1
A circle is the planar locus of all points equidistant from the center O.
Takeaway 2
Diameter is the longest chord: d = 2r.
Takeaway 3
A secant cuts through a circle at two points; a tangent touches at exactly one point.
Takeaway 4
A sector is bounded by two radii and an arc; a segment is bounded by a chord and an arc.
Takeaway 5
The perpendicular line dropped from the center to a chord bisects the chord: AM = MB.
Takeaway 6
Pythagorean relationship for chords: r^2 = d^2 + (c/2)^2.
Takeaway 7
Equal chords in a circle are equidistant from the center.
Takeaway 8
The angle inscribed in a semicircle subtended by a diameter is always 90 degrees.
Takeaway 9
Concentric circles share the identical center but have different radii.
Takeaway 10
Longer chords lie closer to the center than shorter chords.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
A chord of length $24\text{ cm}$ is at a distance of $5\text{ cm}$ from the center of a circle. Find the radius of the circle.
Reveal Answer & Explanation
Answer:

Step 1: The perpendicular from the center bisects the chord:

$$AM = \frac{AB}{2} = \frac{24\text{ cm}}{2} = 12\text{ cm}$$


Step 2: Distance from center $OM = 5\text{ cm}$.
Step 3: In right-angled triangle $\triangle OMA$, by Pythagoras Theorem:

$$OA^2 = OM^2 + AM^2$$


$$r^2 = 5^2 + 12^2 = 25 + 144 = 169$$


$$r = \sqrt{169} = \mathbf{13\text{ cm}}$$

.
The radius of the circle is $13\text{ cm}$.


$r^2 = 5^2 + 12^2 = 25 + 144 = 169 \implies r = 13\text{ cm}$.
2
The radius of a circle is $17\text{ cm}$. Find the length of a chord which is at a distance of $8\text{ cm}$ from the center.
Reveal Answer & Explanation
Answer: Step 1: Let chord be $AB$ and its distance from center $OM = 8\text{ cm}$, radius $r = 17\text{ cm}$.
Step 2: In right $\triangle OMA$:
$$AM^2 = OA^2 - OM^2$$
$$AM^2 = 17^2 - 8^2 = 289 - 64 = 225$$
$$AM = \sqrt{225} = 15\text{ cm}$$
Step 3: Since $OM$ bisects $AB$, the full length of the chord is:
$$AB = 2 \times AM = 2 \times 15 = \mathbf{30\text{ cm}}$$.
$AM = \sqrt{17^2 - 8^2} = \sqrt{225} = 15\text{ cm}$. Chord $AB = 2 \times 15 = 30\text{ cm}$.
3
In a circle of radius $5\text{ cm}$, $AB$ and $CD$ are two parallel chords of length $8\text{ cm}$ and $6\text{ cm}$ respectively. Calculate the distance between the chords if they lie on: (a) opposite sides of the center, (b) the same side of the center.
Reveal Answer & Explanation
Answer:

Step 1: Calculate the distance of each chord from the center $O$:
• Half of chord $AB = \frac{8}{2} = 4\text{ cm}$. Distance $d_1 = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = \mathbf{3\text{ cm}}$.
• Half of chord $CD = \frac{6}{2} = 3\text{ cm}$. Distance $d_2 = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = \mathbf{4\text{ cm}}$.

• (a) Chords on opposite sides of the center:

$$\text{Distance between chords} = d_1 + d_2 = 3 + 4 = \mathbf{7\text{ cm}}$$



• (b) Chords on the same side of the center:

$$\text{Distance between chords} = d_2 - d_1 = 4 - 3 = \mathbf{1\text{ cm}}$$

.


$d_1 = \sqrt{25 - 16} = 3\text{ cm}$, $d_2 = \sqrt{25 - 9} = 4\text{ cm}$. Opposite sides: $3 + 4 = 7\text{ cm}$. Same side: $4 - 3 = 1\text{ cm}$.
4
$AB$ is a diameter of a circle with center $O$. $C$ is a point on the circle such that $AC = 7\text{ cm}$ and $BC = 24\text{ cm}$. Find the radius of the circle.
Reveal Answer & Explanation
Answer: Step 1: By the Semicircle Theorem, the angle subtended by a diameter is a right angle:
$$\angle ACB = 90^{\circ}$$
Step 2: Triangle $\triangle ABC$ is a right-angled triangle with hypotenuse $AB$ (the diameter). By Pythagoras Theorem:
$$AB^2 = AC^2 + BC^2$$
$$AB^2 = 7^2 + 24^2 = 49 + 576 = 625$$
$$AB = \sqrt{625} = 25\text{ cm}$$
Step 3: Calculate the radius:
$$\text{Radius } r = \frac{\text{Diameter } AB}{2} = \frac{25}{2} = \mathbf{12.5\text{ cm}}$$.
$\\angle ACB = 90^{\\circ}$. Diameter $AB = \\sqrt{7^2 + 24^2} = 25\text{ cm}$. Radius $= 25 / 2 = 12.5\text{ cm}$.
5
Differentiate between a Sector of a circle and a Segment of a circle.
Reveal Answer & Explanation
Answer:

• Sector: The region enclosed by two radii and the arc between them (bounded by two line segments and one curved arc, like a pie slice).
• Segment: The region enclosed by a single chord and the arc subtended by that chord (bounded by one straight line segment and one curved arc).


Sector is bounded by 2 radii and an arc; segment is bounded by 1 chord and an arc.
6
Prove that of two unequal chords in a circle, the longer chord is closer to the center.
Reveal Answer & Explanation
Answer: Let the chords be $c_1$ and $c_2$ with distances $d_1$ and $d_2$ from the center in a circle of radius $r$.
By Pythagoras theorem:
$$d_1^2 = r^2 - \left(\frac{c_1}{2}\right)^2 \quad \text{and} \quad d_2^2 = r^2 - \left(\frac{c_2}{2}\right)^2$$
If $c_1 > c_2$, then $\left(\frac{c_1}{2}\right)^2 > \left(\frac{c_2}{2}\right)^2$.
Subtracting a larger positive quantity from $r^2$ makes the result smaller:
$$d_1^2 < d_2^2 \implies \mathbf{d_1 < d_2}$$
Therefore, the longer chord $c_1$ lies closer to the center than the shorter chord $c_2$. Proved!
Since $d^2 = r^2 - (c/2)^2$, as chord $c$ increases, distance $d$ decreases.
7
What is the maximum number of points at which a straight line can intersect a circle? Name the lines.
Reveal Answer & Explanation
Answer:

• $0$ points: An external non-intersecting line.
• $1$ point: A Tangent, which touches the circle at exactly one point of contact.
• $2$ points: A Secant, which cuts through the circle at two distinct boundary points.
• A straight line can never intersect a circle at more than $2$ points because a circle is a degree-2 curve.


At most 2 points (a secant). At 1 point, it is a tangent.
8
Define Concentric Circles and state whether they ever intersect.
Reveal Answer & Explanation
Answer:

• Concentric Circles are two or more circles lying in the same plane that share the exact same central point ($O$), but possess different radii ($r_1 \ne r_2$).
• Because every point on circle 1 is at distance $r_1$ from $O$ and every point on circle 2 is at distance $r_2$ from $O$, they can NEVER intersect or touch each other at any point. The annular space between them forms a circular ring.


Concentric circles share the identical center with different radii. They never intersect.
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