NCERT Class 9 Maths Chapter 9 Circles Notes

Chapter 9 · CBSE · Class IX

⭕

Angle Subtended by a Chord at a Point

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📘 Definition

A chord of a circle can form angles at different points on the circumference. The angle formed at a point on the circle by joining that point to the endpoints of the chord is called the angle subtended by the chord.

Suppose \(\small AB\) is a chord of a circle with centre \(\small O\). Let \(\small P\) be any point on the circumference other than \(\small A\) and \(\small B\). Join \(\small PA\) and \(\small PB\). Then the angle \(\small \angle APB\) is called the angle subtended by chord \(\small AB\) at point \(\small P\).

🗂️ Types / Category

Chord

A line segment joining any two points on a circle.

Arc

A portion of the circumference of a circle.

Minor Arc

The smaller part of the circle cut by a chord.

Major Arc

The larger part of the circle cut by a chord.

Subtended Angle

The angle formed by joining the endpoints of a chord to a point on the circle.

📌 Concept Explanation

Every chord of a circle divides the circle into two arcs. When a point is taken on one of these arcs and connected to the endpoints of the chord, an angle is formed.

This angle depends on the position of the point on the circumference. However, an important property of circles states that:

Angles subtended by the same chord in the same segment of a circle are equal.

Therefore, if points \(\small P\) and \(\small Q\) lie on the same arc of chord \(\small AB\), then:

\[\small \angle APB = \angle AQB \]

This property is one of the most important theorems in coordinate geometry and circle geometry for CBSE board examinations.

🎨 SVG Diagram

Angle Subtended by a chord on a point

🗒️ Relation

Relation Between Central Angle and Angle at Circumference

If a chord subtends an angle at the centre and also at a point on the circle, then the angle at the centre is always twice the angle at the circumference.

\[ \angle AOB = 2\angle APB \]

Here:

\(O\) is the centre of the circle.
\(\angle AOB\) is the angle subtended by chord \(AB\) at the centre.
\(\angle APB\) is the angle subtended by the same chord at point \(P\) on the circumference.

📐 Derivation

Derivation of Central Angle Theorem

Consider a circle with centre \(\small O\). Chord \(\small AB\) subtends:

Angle \(\small \angle AOB\) at the centre
Angle \(\small \angle APB\) at point \(\small P\) on the circle

Join \(\small OP\). Since all radii are equal:

\[\small OA = OP = OB \]

Hence triangles \(\small OAP\) and \(\small OBP\) are isosceles triangles.

By angle sum property and simple angle calculations, we obtain:

\[\small \angle AOB = 2\angle APB \]

This theorem is frequently used in board examinations and HOTS questions.

✏️ Example

Solved Examples

In a circle, chord \(\small AB\) subtends an angle of \(\small 35^\circ\) at a point \(\small P\) on the circle. Find the angle subtended by the same chord at the centre.

Angle at the centre is twice the angle at the circumference.

1

Identify the angle at the circumference.
2

Apply central angle theorem.

Given
\[\small \angle APB = 35^\circ\]
Using Theorem
\[\small \angle AOB = 2\angle APB\]
\[\begin{aligned}\small \angle AOB &= 2 \times 35^\circ\\\angle AOB &= 70^\circ \end{aligned}\]
Therefore, the angle subtended at the centre is \(\small 70^\circ\).

Two angles subtended by chord \(\small AB\) at points \(\small P\) and \(\small Q\) are formed in the same segment of a circle. If \(\small \angle APB = 48^\circ\), find \(\small \angle AQB\).

Angles in the same segment are equal.

\[small \angle APB = \angle AQB \]

\[small \angle AQB = 48^\circ \]

Hence, the required angle is \(\small {48^\circ}\).

⚡ Exam Tip

Always identify whether the angle is formed at the centre or at the circumference.
Remember the theorem: \[\small \text{Angle at centre} = 2 \times \text{Angle at circumference} \]
Angles in the same segment are always equal.
Carefully observe whether points lie on the same arc.
Draw rough figures in geometry proofs for better understanding.

❌ Common Mistakes

Confusing chord with diameter.
Using the theorem incorrectly by doubling the wrong angle.
Forgetting that points must lie on the same segment for equal angles.
Drawing inaccurate diagrams leading to incorrect assumptions.
Ignoring proper notation such as \(\angle APB\).

📋 Case Study

A circular park has two entry gates \(\small A\) and \(\small B\). CCTV cameras are installed at two points \(\small P\) and \(\small Q\) on the boundary of the park. Both cameras observe the gates under equal viewing angles.

If the angle subtended by gates \(\small A\) and \(\small B\) at camera \(\small P\) is \(\small 62^\circ\), answer the following:

Find the angle subtended at camera \(\small Q\).
Find the angle subtended by gates \(\small A\) and \(\small B\) at the centre of the park.

Solution:

Since angles in the same segment are equal:

\[\small \angle AQB = 62^\circ \]

Also:

\[\small \angle AOB = 2 \times 62^\circ \]

\[\small \angle AOB = 124^\circ \]

🌟 Importance

Why This Topic is Important for Board Exams

The concept of angle subtended by a chord forms the foundation of several important theorems in circles.

Questions based on this topic are frequently asked in:

Proof based geometry questions
Case study based CBSE questions
Assertion-reason questions
HOTS problems involving angles
Construction and theorem applications

A strong understanding of this theorem helps students solve advanced geometry problems quickly and accurately.

⭕

Relationship between the Size of a Chord and the Angle Subtended by it at the Centre

🗺️ Overview

In a circle, different chords subtend different angles at the centre. The size of the chord directly affects the size of the central angle formed by it.

The larger the chord, the larger is the angle subtended by it at the centre of the circle. Similarly, the smaller the chord, the smaller is the angle subtended at the centre.

💡 Concept

Consider a circle with centre \(\small O\). Let \(\small AB\) and \(\small CD\) be two chords such that:

\[\small AB > CD \]

Then the angle subtended by chord \(\small AB\) at the centre will also be greater.

\[\small \angle AOB > \angle COD \]

Therefore:

Equal chords subtend equal angles at the centre.
Larger chords subtend larger angles at the centre.
Smaller chords subtend smaller angles at the centre.

🗒️ Relation

Relation with Arc of the Circle

This relationship is closely connected with the arc formed by the chord.

A longer chord forms a larger arc.
A shorter chord forms a smaller arc.
The angle at the centre depends on the size of the arc intercepted by the chord.

Hence, larger arcs correspond to larger central angles.

📌 Note

The converse is also true.

If two chords subtend equal angles at the centre, then the chords are equal in length.

\[\small \angle AOB = \angle COD \Rightarrow AB = CD \]

🔢 Formula

Formula Summary

For chords in the same circle or congruent circles:

Larger chord \(\small \Rightarrow \) Larger central angle
Smaller chord \(\small \Rightarrow \) Smaller central angle
Equal chords \(\small \Rightarrow \) Equal central angles

✏️ Example

Solved Example

In a circle with centre \(\small O\), chord \(\small AB\) is longer than chord \(\small CD\). Compare \(\small \angle AOB\) and \(\small \angle COD\).

Larger chords subtend larger angles at the centre.

Since:

\[\small AB > CD \]

Therefore:

\[\small \angle AOB > \angle COD \]

Hence, the angle subtended by chord \(\small AB\) at the centre is greater.

⚡ Exam Tip

❌ Common Mistakes

Assuming equal arcs always mean equal radii.
Comparing chords from different circles without checking radii.
Confusing angle at the centre with angle at the circumference.
Using visual estimation instead of theorem-based reasoning.

⭕

Theorem 1: Equal Chords of a Circle Subtend Equal Angles at the Centre

📌 Note

Theorem Statement

🔬 Proof

Given:

\(\small AB\) and \(\small CD\) are two equal chords of a circle with centre \(\small O\).

\[\small AB = CD \]

To Prove: \[\small \angle AOB = \angle COD\]

Construction:

Join:

\(\small OA\)
\(\small OB\)
\(\small OC\)
\(\small OD\)

These are radii of the same circle.

🔬 Proof

Consider triangles \(\small \triangle AOB\) and \(\small \triangle COD\).
Since all radii of a circle are equal:

\[\small OA = OC\] \[\small OB = OD\] \[\small AB = CD\]
Hence

\[\small \triangle AOB \cong \triangle COD\]
by SSS congruence rule.
Therefore, corresponding angles are equal:
\[\small \angle AOB = \angle COD\]

🌟 Importance

Why this Theorem is Important

This theorem establishes a direct relationship between:

Length of a chord
Angle subtended by the chord at the centre

It forms the foundation for many advanced circle theorems used in:

Geometry proofs
Circle constructions
Case study questions
CBSE competency-based questions
Coordinate geometry applications

📌 Converse of theorem

The converse of this theorem is also true.

If two chords subtend equal angles at the centre, then the chords are equal.

\[\small \begin{aligned} \angle AOB &= \angle COD\\ \Rightarrow AB &= CD \end{aligned} \]

💡 Conceptual Understanding

A larger chord stretches farther across the circle, therefore it creates a larger opening at the centre.

Similarly:

Equal chords create equal openings at the centre.
Unequal chords create unequal central angles.

✏️ Example

In a circle with centre \(\small O\), chords \(\small AB\) and \(\small CD\) are equal. If: \[\small \angle AOB = 75^\circ \] Find \(\small \angle COD\).

Equal chords subtend equal angles at the centre.

Since:

\[\small AB = CD \]

Therefore:

\[\small \angle AOB = \angle COD \]

Given:

\[\small \angle AOB = 75^\circ \]

Hence:

\[\small \angle COD = 75^\circ \]

⚡ Exam Tip

Always check whether the chords belong to the same circle.
Mention “Radii of the same circle are equal” explicitly in proofs.
Write the congruence rule properly while proving the theorem.
Use CPCT only after proving triangle congruence.
Draw neat labelled diagrams in board examinations.

❌ Common Mistakes

Using SAS instead of SSS without justification.
Forgetting to state that \(OA, OB, OC,\) and \(OD\) are radii.
Writing CPCT before proving congruence.
Confusing angle at the centre with angle at the circumference.
Comparing chords from different circles of unequal radii.

📋 Case Study

CBSE HOTS Question

In a circular stadium, two equal banners are fixed along chords \(\small AB\) and \(\small CD\). The banners subtend angles \(\small \angle AOB\) and \(\small \angle COD\) at the centre respectively.

If:

\[\small \angle AOB = 3x + 10^\circ \]

and

\[\small \angle COD = 5x - 30^\circ \]

Find the value of \(\small x\).

Solution:

Since equal chords subtend equal angles:

\[\small 3x + 10 = 5x - 30 \]

\[\small 40 = 2x \]

\[\small x = 20 \]

⭕

Theorem 2: If Equal Angles are Subtended at the Centre, then the Chords are Equal

📘 Theorem Statement

🔬 Proof

Given:

In a circle with centre \(\small O\):

\[\small \angle AOB = \angle COD \]

To Prove: \[\small AB = CD\]

Construction:

Join:

\(OA\)
\(OB\)
\(OC\)
\(OD\)

These are radii of the same circle.

🔬 Proof

Consider triangles \(\small \triangle AOB\) and \(\small \triangle COD\).
Since all radii of the same circle are equal:

\[\small OA = OC\] \[\small OB = OD\]
Also

\[\small \angle AOB = \angle COD\]
Given
Therefore, in triangles \(\small\triangle AOB\) and \(\small\triangle COD\):

\[\small \begin{aligned} OA &= OC\\ \angle AOB &= \angle COD\\ OB &= OD \end{aligned} \]
Hence:

\[\small \triangle AOB \cong \triangle COD\]
by SAS congruence rule.
Therefore, corresponding sides are equal:

\[\small AB = CD\]
by CPCT.
Hence Proved.

📌 Note

This theorem is the converse of the previous theorem:

Equal chords subtend equal angles at the centre.
Equal central angles subtend equal chords.

Thus, chord length and central angle are directly related.

💡 Concept

Key Concept

🔢 Formula

\[\small \angle AOB = \angle COD \Rightarrow AB = CD \]
Equal central angles \(\small \Rightarrow \) Equal chords
Equal chords \(\small \Rightarrow \) Equal central angles

🗒️ Example

In a circle with centre \(\small O\), chords \(\small AB\) and \(\small CD\) subtend equal angles at the centre. If: \[\small \angle AOB = \angle COD \] and \[\small AB = 8\text{ cm} \] Find the length of chord \(\small CD\).

Equal angles at the centre subtend equal chords.

Since:

\[\small \angle AOB = \angle COD \]

Therefore:

\[\small AB = CD \]

Given:

\small \[ AB = 8\text{ cm} \]

Hence:

\[\small CD = 8\text{ cm} \]

⚡ Exam Tip

Always mention the congruence rule clearly.
Use SAS only after writing all equal sides and included angle.
Mention “Radii of the same circle are equal”.
Write CPCT only after proving congruence.
Draw labelled figures neatly in geometry proofs.

❌ Common Mistakes

Using SSS instead of SAS in this proof.
Forgetting that the equal angle must be included between equal sides.
Writing congruence without justification.
Confusing chord equality with radius equality.
Using the theorem in different circles without checking radius conditions.

📋 Case Study

A circular amusement park has two equal pathways represented by chords \(\small AB\) and \(\small CD\). The pathways subtend equal angles at the centre of the park.

If:

\[\small AB = 3x + 2 \]

and:

\[\small CD = 5x - 8 \]

Find the value of \(\small x\).

Solution:

Since equal central angles subtend equal chords:

\[\small AB = CD \]

Therefore:

\[\small 3x + 2 = 5x - 8 \]

\[\small 10 = 2x \]

\[\small x = 5 \]

⭕

Theorem 3: The Perpendicular from the Centre of a Circle to a Chord Bisects the Chord

📘 Theorem Statement

🔬 Proof

Given:

In a circle with centre \(\small O\), \(\small AB\) is a chord and \(\small OM\) is perpendicular to \(\small AB\).

\[\small \angle OMA = \angle OMB = 90^\circ \]

To Prove: \[\small MA = MB\]

Construction:

Join:

\(OA\)
\(OB\)

Since \(OA\) and \(OB\) are radii of the circle, they are equal.

🔬 Proof

Consider triangles \(\triangle OAM\) and \(\triangle OBM\).
Since all radii of the same circle are equal:

\[\small OA = OB\]
Also,

\[\small OM = OM\]
because it is the common side.
\[\small \angle OMA = \angle OMB = 90^\circ\]
Given
Therefore, in right triangles \(\small \triangle OAM\) and \(\small \triangle OBM\):

\[\small \begin{aligned} OA ^= OB\\ OM ^= OM\\ \angle OMA ^= \angle OMB = 90^\circ \end{aligned}\]
Hence

\[\small \triangle OAM \cong \triangle OBM\]
by RHS congruence rule.
\[\small MA = MB\]
Hence Proved

💡 Conceptual Understanding

A perpendicular from the centre reaches the chord exactly at its midpoint.

This means:

The chord gets divided into two equal parts.
The perpendicular acts as a line of symmetry for the chord.
The two triangles formed on either side become congruent.

📌 Converse of Theorem

The converse of this theorem is also true.

If a line drawn from the centre bisects a chord, then it is perpendicular to the chord.

\[\small MA = MB \Rightarrow OM \perp AB \]

🔢 Formula

Formula Summary

Perpendicular from centre \(\small \Rightarrow \) Chord bisected
\[\small OM \perp AB \Rightarrow MA = MB \]
Converse: \[\small MA = MB \Rightarrow OM \perp AB \]

✏️ Example

Solved Example

In a circle with centre \(\small O\), chord \(\small AB = 16\text{ cm}\). If \(\small OM \perp AB\), find \(\small MA\).

Perpendicular from the centre bisects the chord.

Since:

\[\small OM \perp AB \]

Therefore:

\[\small MA = MB \]

Also:

\[\small AB = MA + MB \]

Since:

\[\small MA = MB \]

Therefore:

\[\small MA = \frac{AB}{2} \]

\[\small MA = \frac{16}{2} \]

\[\small MA = 8\text{ cm} \]

⚡ Exam Tip

Always identify the right angle carefully in the figure.
Mention RHS congruence rule correctly.
Use “Radii of the same circle are equal” explicitly.
Remember that the theorem works only when the perpendicular is drawn from the centre.
Draw neat perpendicular marks in geometry diagrams.

❌ Common Mistakes

Forgetting that \(\small OM\) must pass through the centre.
Using SAS instead of RHS in the proof.
Assuming every line through the centre bisects a chord.
Confusing midpoint theorem with circle theorem.
Writing CPCT before proving congruence.

📋 Case Study

A circular fountain has a straight decorative rod represented by chord \(\small AB\). A support pipe from the centre \(\small O\) touches the rod at point \(\small M\) such that:

\[\small OM \perp AB \]

If:

\[\small AB = 24\text{ m} \]

find the lengths of \(\small MA\) and \(\small MB\).

Solution:

Since perpendicular from the centre bisects the chord:

\[\small MA = MB \]

Also:

\[\small AB = MA + MB \]

Therefore:

\[\small MA = MB = \frac{24}{2} \]

\[\small \boxed{MA = MB = 12\text{ m}} \]

⭕

Theorem 4: The Line Drawn Through the Centre to Bisect a Chord is Perpendicular to the Chord

📘 Theorem Statement

🔬 Proof

Given:

In a circle with centre \(\small O\), chord \(AB\) is bisected at point \(\small M\).

\[\small MA = MB \]

To Prove: \[\small \angle OMA = \angle OMB = 90^\circ\] Therefore: \[\small OM \perp AB\]

Construction:

Join:

\(\small OA\)
\(\small OB\)

Since \(\small OA\) and \(\small OB\) are radii of the same circle, they are equal.

🔬 Proof

Consider triangles \(\small \triangle OAM\) and \(\small \triangle OBM\).
Since all radii of the same circle are equal:

\[\small OA = OB\]
Also

\[\small OM = OM\]
Common Side
\[\small MA = MB\]
Given
Therefore, in triangles \(\small\triangle OAM\) and \(\small\triangle OBM\):

\[\small\begin{aligned} OA &= OB\\ OM &= OM\\ MA &= MB \end{aligned} \]
Hence,

\[\small \triangle OAM \cong \triangle OBM\]
by SSS congruence rule.
Therefore:

\[\small \angle OMA = \angle OMB\]
by CPCT.
Since points \(\small A\), \(\small M\), and \(\small B\) lie on the same straight line, \(\small \angle OMA\) and \(\small \angle OMB\) form a linear pair.
Therefore:

\[\small \angle OMA + \angle OMB = 180^\circ]\]
Since,

\[\small \angle OMA = \angle OMB\]
Therefore:

\[\small \begin{aligned}2\angle OMA &= 180^\circ\\2\angle OMA &= 180^\circ\\\angle OMA &= 90^\circ \end{aligned}\]
Similarly:

\[\small \angle OMB = 90^\circ\]
Hence:

\[\small \angle OMA = \angle OMB = 90^\circ\]
Therefore:

\[\small OM \perp AB\]
Hence Proved

💡 Conceptual Understanding

If a line from the centre divides a chord into two equal parts, then the line must meet the chord at right angles.

This theorem is the converse of the previous theorem:

Perpendicular from centre \(\small \Rightarrow \) Chord bisected
Chord bisected by line through centre \(\small \Rightarrow \) Perpendicular formed

📌 Important Note

🔢 Formula

Formula Summary

\[\small MA = MB \Rightarrow OM \perp AB \]
Line through centre bisecting chord \(\small \Rightarrow \) Right angle formed
Converse of Theorem 3

✏️ Example

Solved Example

In a circle with centre \(\small O\), point \(\small M\) is the midpoint of chord \(\small AB\). Show that \(\small OM\) is perpendicular to \(\small AB\).

A line through the centre bisecting a chord is perpendicular to the chord.

Since \(\small M\) is the midpoint of chord \(\small AB\):

\[\small MA = MB \]

Also, \(\small O\) is the centre of the circle.

Therefore, by the theorem:

\[\small OM \perp AB \]

⚡ Exam Tip

Mention that \(\small OA\) and \(\small OB\) are radii.
Use SSS congruence correctly in the proof.
Write the linear pair property explicitly.
Do not skip the CPCT step.
Always mention that the line passes through the centre.

❌ Common Mistakes

Forgetting the condition that the line passes through the centre.
Using RHS instead of SSS in this theorem.
Forgetting to prove equal angles before applying linear pair.
Assuming every midpoint line is perpendicular automatically.
Missing the final conclusion: \[\small OM \perp AB \]

📋 Case Study

A circular playground has a rope represented by chord \(\small AB\). Point \(\small M\) is the midpoint of the rope. The centre of the playground is \(\small O\).

If:

\[\small MA = MB = 7\text{ m} \]

determine the angle formed between \(\small OM\) and chord \(\small AB\).

Solution:

Since:

\[\small MA = MB \]

and the line passes through the centre, by the theorem:

\[\small OM \perp AB \]

Therefore:

\[\small \boxed{\angle OMA = 90^\circ} \]

⭕

Theorem 5: Equal Chords of a Circle are Equidistant from the Centre

📘 Theorem Statement

📌 Note

Meaning of Equidistant

Two objects are said to be equidistant from a point if they are at the same perpendicular distance from that point.

In this theorem:

The point is the centre of the circle.
The objects are the chords.
Distance is measured using perpendiculars drawn from the centre to the chords.

🗒️ Svg

🔬 Proof

Given:

In a circle with centre \(\small O\), \(\small AB\) and \(\small CD\) are equal chords.

\[\small AB = CD \]

Perpendiculars:

\[\small OM \perp AB \]

and:

\[\small ON \perp CD \]

To Prove: \[\small OM = ON\]

🔬 Proof

Since perpendicular from the centre bisects the chord:

\[\small AM = MB\] \[\small CN = ND\]
\[\small AB = CD\]
Therefore:

\[\small AM = CN\]
Consider right triangles \(\triangle OMA\) and \(\triangle ONC\).
Since:

\[\small OA = OC\]
Also:

\[\small AM = CN\] and \[\small \angle OMA = \angle ONC = 90^\circ\]
Therefore:

\triangle OMA \cong \triangle ONC
Hence:

\[\small OM = ON\]
Therefore, equal chords are equidistant from the centre.

💡 Concept

🔢 Formula

Formula Summary

🗒️ Solved Example

Solved Example

Two equal chords of a circle are at distances \(\small x + 3\) cm and \(\small 2x - 5\) cm from the centre. Find \(\small x\).

Equal chords are equidistant from the centre.

\[\small x + 3 = 2x - 5 \]

\[\small x = 8 \]

Hence:

\[\small x = 8 \]

⚡ Exam Tip

Distance from the centre is always measured perpendicular to the chord.
Do not measure slant distance as chord distance.
Use RHS congruence carefully in proofs.
Mention midpoint property before applying congruence.

❌ Common Mistakes

Measuring distance obliquely instead of perpendicular distance.
Forgetting that perpendicular bisects the chord.
Applying theorem to unequal circles.
Using SAS instead of RHS unnecessarily.

⭕

Theorem 6: Chords Equidistant from the Centre are Equal in Length

📘 Theorem Statement

🗒️ Svg

🔬 Proof

Given: In a circle with centre \(\small O\), perpendicular distances of chords \(\small AB\) and \(\small CD\) from the centre are equal. \[\small OM = ON\]

To Prove: \[\small AB = CD\]

🔬 Proof

Since perpendicular from the centre bisects the chord:

\[\small AM = MB\] and \[\small CN = ND\]
Consider right triangles \(\small \triangle OMA\) and \(\small \triangle ONC\).
Since:

\[\small OA = OC\]
radii od sme circle
Also:

\[\small OM = ON\] and \[\small \angle OMA = \angle ONC = 90^\circ\]
Therefore:

\[\small \triangle OMA \cong \triangle ONC\]
by RHS congruence rule.
Hence:

\[\small AM = CN\]
Since

\[\small AB = 2AM\] and \[\small CD = 2CN\]
Therefore:

\[\small AB = CD\]
Hence proved.

💡 Conceptual Understanding

🔢 Formula

Formula Summary

✏️ Example

Solved Example

Two chords of a circle are at equal distances from the centre. If one chord measures \(\small 12\text{ cm}\), find the length of the other chord.

Since the chords are equidistant from the centre, they are equal in length.

Therefore:

\[\small 12\text{ cm} \]

⚡ Exam Tip

Distance from centre must always be perpendicular.
The theorem applies only within the same circle or congruent circles.
Use midpoint property carefully while proving equality of chords.
Learn converse relations of all circle theorems together.

❌ Common Mistakes

Confusing radius with perpendicular distance.
Forgetting that equal distance means perpendicular distance only.
Applying theorem to arcs instead of chords.
Missing the relation: \[\small AB = 2AM \]

📋 Case Study

In a circular garden, two benches are placed along chords \(\small AB\) and \(\small CD\). Both benches are at equal perpendicular distances from the centre.

If:

\[\small AB = 18\text{ m} \]

find the length of chord \(\small CD\).

Solution:

Since the chords are equidistant from the centre:

\[\small AB = CD \]

Therefore:

\[\small CD = 18\text{ m} \]

⭕

Theorem 7: Angle Subtended by an Arc at the Centre is Double the Angle Subtended at the Remaining Part of the Circle

📘 Definition

🎨 SVG Diagram

🔬 Proof

Given:

Arc \(\small PQ\) of a circle subtends:

\(\small \angle POQ\) at the centre \(\small O\)
\(\small \angle PAQ\) at a point \(\small A\) on the remaining part of the circle

To Prove: \[\small\angle POQ = 2\angle PAQ\]

Construction:

Join:

\(AO\)

Extend line \(AO\) to meet the circle at point \(B\).

🔬 Proof

In triangle \(\small\triangle AOQ\), \(\small\angle BOQ\) is an exterior angle.
Therefore:

\[\small \angle BOQ = \angle OAQ + \angle AQO \tag{1}\]
Similarly, in triangle \(\triangle AOP\):

\[\small\angle POB = \angle OAP + \angle OPA\tag{2}\]
Since \(\small OA = OQ\), triangle \(\small \triangle AOQ\) is isosceles.
Therefore:

\[\small \angle OAQ = \angle AQO\]
Also, since \(\small OA = OP\), triangle \(\small \triangle AOP\) is isosceles, Therefore:

\[\small \angle OAP = \angle OPA\]
Adding equations (1) and (2):

\[\small\angle BOQ + \angle POB=\angle OAQ + \angle AQO + \angle OAP + \angle OPA\]
Since equal angles are substituted:

\[\small \angle POQ=2\angle OAQ + 2\angle OAP\]
Taking common factor:

\[\small \angle POQ=2(\angle OAQ + \angle OAP)\]
Since:

\[\small \angle OAQ + \angle OAP = \angle PAQ\]
Therefore:

\[\small \angle POQ = 2\angle PAQ\]
Hence Proved

💡 Conceptual Understanding

The central angle always spreads wider than the angle formed at the circumference.

This is because the centre is closer to the intercepted arc, while the point on the circle lies farther away.

Therefore:

Central angle \(\small =\) Twice the angle at the circumference.

🗒️ Important Result

If an angle subtended by an arc at the centre is:

\[\small \theta \]

then the angle subtended at the circumference by the same arc is:

\[\small \frac{\theta}{2} \]

🔢 Formula

Formula Summary

\(\small \angle \text{ at centre} = 2 \times \angle \text{ at circumference} \)
\(\small \angle POQ = 2\angle PAQ \)
\(\small \angle PAQ = \frac{1}{2}\angle POQ \)

✏️ Example

Solved Example

An arc subtends an angle of \(\small 120^\circ\) at the centre of a circle. Find the angle subtended by the same arc at a point on the remaining part of the circle.

Angle at the centre is twice the angle at the circumference.

Given:

\[\small \angle POQ = 120^\circ \]

Using theorem:

\[\small \angle PAQ = \frac{1}{2}\angle POQ \]

\[\small \angle PAQ = \frac{120^\circ}{2} \]

\[\small \angle PAQ = 60^\circ \]

📋 Case Study

In a circle with centre \(\small O\), an arc subtends an angle:

\[\small \angle POQ = 5x + 10^\circ \]

at the centre and:

\[\small \angle PAQ = 3x - 10^\circ \]

at a point on the remaining part of the circle. Find \(\small x\).

Solution:

Using theorem:

\[\small \angle POQ = 2\angle PAQ \]

Therefore:

\[\small 5x + 10 = 2(3x - 10) \]

\[\small 5x + 10 = 6x - 20 \]

\[\small x = 30 \]

Hence:

\[\small x = 30 \]

⚡ Exam Tip

Learn the theorem statement exactly as given in NCERT.
Remember: Centre angle is always larger.
Use isosceles triangle properties carefully in proofs.
Mention exterior angle property explicitly.
Draw properly labelled figures in proofs.

❌ Common Mistakes

Reversing the theorem incorrectly.
Forgetting to divide the central angle by 2.
Using wrong arcs while identifying angles.
Skipping construction in theorem proof.
Missing isosceles triangle angle relations.

⭕

Theorem 8: Angles in the Same Segment of a Circle are Equal

📘 Theorem Statement

📌 Meaning of Same Segment

A chord divides a circle into two regions called segments.

Minor segment
Major segment

If two points lie on the same segment of a chord, then the angles formed by joining these points to the endpoints of the chord are equal.

🎨 SVG Diagram

🔬 Proof

Given:

In a circle, chord \(\small AB\) subtends:

\(\small \angle APB\) at point \(P\)
\(\small \angle AQB\) at point \(Q\)

where points \(\small P\) and \(\small Q\) lie on the same segment of the circle.

To Prove: \[\small \angle APB = \angle AQB\]

🔬 Proof

Let \(\small O\) be the centre of the circle.
Chord \(\small AB\) subtends:
- \(\small \angle AOB\) at the centre
- \(\small \angle APB\) and \(\angle AQB\) at points on the circumference
By Theorem 7:

\[\small \angle AOB = 2\angle APB \tag{1}\]
Also:

\[\small \angle AOB = 2\angle AQB\tag{2}\]
From equations (1) and (2):

\[\small 2\angle APB = 2\angle AQB\]
Dividing both sides by \(2\):

\[\small \angle APB = \angle AQB\]
Hence proved.

💡 Conceptual Understanding

When the same chord is viewed from different points on the same segment, the opening formed remains unchanged.

Therefore:

All angles standing on the same chord and lying in the same segment are equal.

📌 Important Note

🔢 Formula

Formula Summary

Angles in the same segment are equal.
\[ \angle APB = \angle AQB \]
Same chord \( \Rightarrow \) Equal angles in same segment

✏️ Example

In a circle, chord \(\small AB\) subtends angles: \[\small \angle APB = 65^\circ\] and \(\small \angle AQB\) in the same segment. Find \(\small \angle AQB\).

Angles in the same segment are equal.

Since both angles lie in the same segment: \[\small \angle APB = \angle AQB\] Therefore: \[\small \angle AQB = 65^\circ\]

📋 Case Study

In a circle, chord \(\small AB\) subtends:

\[\small \angle APB = 3x + 5^\circ \]

and:

\[\small \angle AQB = 5x - 15^\circ \]

at points \(\small P\) and \(\small Q\) in the same segment. Find the value of \(\small x\).

Solution:

Since angles in the same segment are equal:

\[\small 3x + 5 = 5x - 15 \]

\[\small 20 = 2x \]

\[\small x = 10 \]

Hence:

\[\small x = 10 \]

⚡ Exam Tip

Identify the chord carefully before applying the theorem.
Ensure the points lie on the same segment.
Mention Theorem 7 if proving this theorem.
Draw proper labelled figures in geometry proofs.
Learn converse and related theorems together for better understanding.

❌ Common Mistakes

Applying the theorem for points on opposite segments.
Confusing central angle with angle in the segment.
Using different chords accidentally.
Forgetting that the same chord must subtend both angles.
Writing equal angles without justification.

🗒️ Important Result Derived From This Theorem

Angles subtended by a diameter at the circumference are always right angles.

This is because the angle subtended by a diameter at the centre is:

\[\small 180^\circ \]

Therefore, the angle at the circumference becomes:

\[\small \frac{180^\circ}{2} = 90^\circ \]

⭕

Theorem 9: Equal Angles Subtended by a Line Segment Imply Concyclic Points

📘 Theorem Statement

📌 Meaning of Concyclic Points

🎨 SVG Diagram

🔬 Proof

Given:

A line segment \(\small AB\) subtends equal angles at points \(\small P\) and \(\small Q\) lying on the same side of line \(\small AB\).

\[\small \angle APB = \angle AQB \]

To Prove: Points \(\small A\), \(\small B\), \(\small P\), and \(\small Q\) are concyclic.

🔬 Proof

We know that equal angles standing on the same chord lie in the same segment of a circle.
Since:

\[\small \angle APB = \angle AQB\]
both angles subtend the same chord \(\small AB\).
Therefore, points \(\small P\) and \(\small Q\) lie in the same segment corresponding to chord \(\small AB\).
Hence:

\[\small A,\ B,\ P,\ Q\] lie on the same circle.
Therefore:

\[\small \text{Points } A,\ B,\ P,\ Q \text{ are concyclic}\]
Hence proved.

💡 Conceptual Understanding

If two equal angles are formed by looking at the same line segment from different points, then those points naturally lie on a common circular path.

This theorem is the converse of the theorem:

Angles in the same segment are equal.

🗒️ Important Result Derived From This Theorem

This theorem is extremely useful for proving that four points are concyclic in geometry problems.

It is frequently used in:

Olympiad geometry
CBSE competency-based questions
Coordinate geometry proofs
Cyclic quadrilateral problems

🔢 Formula

Formula Summary

Equal angles standing on same line segment:

\[\small \angle APB = \angle AQB \]

implies:

\[\small A,\ B,\ P,\ Q \text{ are concyclic} \]

✏️ Example

In a figure, points \(\small P\) and \(\small Q\) lie on the same side of line segment \(\small AB\) such that:\[\small \angle APB = \angle AQB = 50^\circ\] Show that points \(A\), \(B\), \(P\), and \(Q\) are concyclic.

Since:

\[\small \angle APB = \angle AQB \]

both angles subtend the same line segment \(\small AB\).

Therefore, by the converse of angles in the same segment theorem:

\[\small A,\ B,\ P,\ Q \text{ are concyclic} \]

📋 Case Study

In a geometry figure, points \(\small P\) and \(\small Q\) satisfy:

\[\small \angle APB = 3x + 15^\circ \]

and:

\[\small \angle AQB = 5x - 25^\circ \]

If points \(\small A\), \(\small B\), \(\small P\), and \(\small Q\) are concyclic, find the value of \(\small x\).

Solution:

Since points are concyclic:

\[\small \angle APB = \angle AQB \]

Therefore:

\[\small 3x + 15 = 5x - 25 \]

\[\small 40 = 2x \]

\[\small x = 20 \]

⚡ Exam Tip

Learn the meaning of “concyclic” properly.
Equal angles subtending same segment is a strong clue for cyclic figures.
Always check whether points lie on the same side of the line segment.
Use this theorem frequently in proof-based questions.
Draw rough circular figures for better visualization.

❌ Common Mistakes

Forgetting the condition “same side of the line segment”.
Confusing equal angles with supplementary angles.
Writing points are concyclic without theorem justification.
Applying the theorem for unequal subtended chords.

⭕

Theorem 10: Opposite Angles of a Cyclic Quadrilateral are Supplementary

📘 Theorem Statement

🎨 SVG Diagram

🔬 Proof

Given: \(\small ABCD\) is a cyclic quadrilateral.

To Prove: \[\small \angle A + \angle C = 180^\circ\] and \[\small \angle B + \angle D = 180^\circ\]

🔬 Proof

Let \(\small O\) be the centre of the circle.
Angle \(\small \angle A\) subtends arc \(\small BCD\).
Therefore, the angle subtended by the same arc at the centre is:
\[\small \angle AOB = 2\angle A\]
Similarly, angle \(\angle C\) subtends arc \(BAD\).
Therefore, the angle subtended by the same arc at the centre is:
\[\small \angle COD = 2\angle C\]
Since \(\small AOB\) and \(\small COD\) are angles around point \(\small O\):
\[\small \angle AOB + \angle COD = 360^\circ\]
Substituting the values of \(\small \angle AOB\) and \(\small \angle COD\):
\[\small 2\angle A + 2\angle C = 360^\circ\]
Dividing both sides by \(2\):
\[\small \angle A + \angle C = 180^\circ\]
Similarly, we can prove that:
\[\small \angle B + \angle D = 180^\circ\]
Hence proved.

💡 Conceptual Understanding

🔢 Formula

Formula Summary

For cyclic quadrilateral \(\small ABCD\):

\[\small \angle A + \angle C = 180^\circ \]

\[\small \angle B + \angle D = 180^\circ \]

✏️ Example

Solved Example

In cyclic quadrilateral \(\small ABCD\): \[\small \angle A = 75^\circ \] Find \(\small \angle C\).

Opposite angles of a cyclic quadrilateral are supplementary.

Therefore:

\[\small \angle A + \angle C = 180^\circ \]

\[\small 75^\circ + \angle C = 180^\circ \]

\[\small \angle C = 105^\circ \]

Hence:

\[\small \angle C = 105^\circ \]

📋 Case Study

In cyclic quadrilateral \(\small ABCD\):

\[\small \angle A = 4x + 10^\circ \]

and:

\[\small \angle C = 2x + 50^\circ \]

Find the value of \(\small x\).

Solution:

Since opposite angles are supplementary:

\[\small (4x + 10) + (2x + 50) = 180 \]

\[\small 6x + 60 = 180 \]

\[\small 6x = 120 \]

\[\small x = 20 \]

⚡ Exam Tip

Opposite angles are supplementary, not equal.
Use this theorem frequently in proof-based questions.
Check carefully whether the quadrilateral is cyclic.
Learn converse theorem together for better understanding.
Draw proper cyclic figures in exams.

❌ Common Mistakes

Adding adjacent angles instead of opposite angles.
Assuming all quadrilaterals are cyclic.
Writing opposite angles equal instead of supplementary.
Forgetting the \(180^\circ\) relation.

⭕

Theorem 11: If Opposite Angles of a Quadrilateral are Supplementary, then the Quadrilateral is Cyclic

📘 Theorem Statement

🎨 SVG Diagram

🔬 Proof

Given:

In quadrilateral \(\small ABCD\):

\[\small \angle A + \angle C = 180^\circ \]

To Prove: Quadrilateral \(\small ABCD\) is cyclic.

🔬 Proof

Draw a circle with \(\small AB\) as a chord.
Let the circle intersect the extension of \(\small AB\) at point \(\small E\).
Join points \(\small E\) and \(\small C\).
Since \(\small ABE\) is a straight line:
\[\small \angle ABE = 180^\circ\]
Angle \(\small ABE\) subtends arc \(\small AE\).
Therefore, the angle subtended by the same arc at point \(\small C\) is:
\[\small \angle ACE = \angle ABE = 180^\circ - \angle A\]
Since \(\small ACE\) is a straight line:
\[\small \angle ACE + \angle ACD = 180^\circ\]
Substituting the value of \(\small \angle ACE\):
\[\small (180^\circ - \angle A) + \angle ACD = 180^\circ\]
Rearranging: \[\small \angle A + \angle ACD = 180^\circ\]
Since we are given that:
\[\small \angle A + \angle C = 180^\circ\]
Comparing the two equations: \[\small \angle C = \angle ACD\]
Therefore, point \(D\) also lies on the circle.
Hence, points \(A\), \(B\), \(C\), and \(D\) are concyclic.
Therefore, quadrilateral \(ABCD\) is cyclic.
Hence proved.

💡 Cyclic Quadrilateral

🔢 Formula Summary

If:

\[\small \angle A + \angle C = 180^\circ \]

then:

\[\small ABCD \text{ is cyclic} \]

🗒️ Example

Solved Example

In quadrilateral \(\small ABCD\): \[\small \angle A = 95^\circ \] and: \[\small \angle C = 85^\circ \] Show that the quadrilateral is cyclic.

Adding opposite angles:

\[\small 95^\circ + 85^\circ = 180^\circ \]

Therefore:

\[\small \angle A + \angle C = 180^\circ \]

Hence:

\[\small ABCD \text{ is cyclic} \]

📋 Case Study

In quadrilateral \(\small PQRS\):

\[\small \angle P = 3x + 20^\circ \]

and:

\[\small \angle R = 5x \]

If \(\small PQRS\) is cyclic, find \(\small x\).

Solution:

Since opposite angles of a cyclic quadrilateral are supplementary:

\[\small (3x + 20) + 5x = 180 \]

\[\small 8x + 20 = 180 \]

\[\small 8x = 160 \]

\[\small x = 20 \]

⚡ Exam Tip

This theorem is the converse of Theorem 10.
Check opposite angles carefully before concluding cyclicity.
Supplementary opposite angles are a strong clue in proofs.
Use this theorem in geometry constructions and proofs.
Remember:
Converse theorems are frequently asked in CBSE exams.

❌ Common Mistakes

Using adjacent angles instead of opposite angles.
Forgetting to verify the sum equals \(180^\circ\).
Declaring cyclicity without theorem justification.
Confusing converse with original theorem.

⭕

Important Points and Key Results of Circles

⚡ Quick Revision

A circle is the collection of all points in a plane which are equidistant from a fixed point called the centre.
The fixed distance from the centre to any point on the circle is called the radius.
A chord is a line segment joining any two points on a circle.
A diameter is the longest chord of a circle and passes through the centre.
Equal chords of the same circle or congruent circles subtend equal angles at the centre.
If the angles subtended by two chords at the centre are equal, then the chords are equal.
The larger the chord, the larger is the angle subtended by it at the centre.
The perpendicular drawn from the centre of a circle to a chord bisects the chord.
The line drawn through the centre to bisect a chord is perpendicular to the chord.
Equal chords of a circle are equidistant from the centre.
Chords equidistant from the centre are equal in length.
Congruent arcs of a circle subtend equal angles at the centre.
Equal chords correspond to equal arcs and conversely equal arcs correspond to equal chords.
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
The angle at the centre is always twice the angle at the circumference standing on the same arc.
Angles in the same segment of a circle are equal.
Angle subtended by a diameter at the circumference is always: \[\small 90^\circ \]
The angle in a semicircle is always a right angle.
If a line segment subtends equal angles at two points on the same side, then the four points are concyclic.
A quadrilateral whose all vertices lie on a circle is called a cyclic quadrilateral.
Opposite angles of a cyclic quadrilateral are supplementary.
For cyclic quadrilateral \(ABCD\): \[\small \angle A + \angle C = 180^\circ \] and: \[\small \angle B + \angle D = 180^\circ \]
If the sum of a pair of opposite angles of a quadrilateral is \(\small 180^\circ\), then the quadrilateral is cyclic.
The centre of a circle lies on the perpendicular bisector of every chord.
Equal arcs subtend equal angles at the centre as well as at the circumference.
In the same circle, equal central angles correspond to equal arcs and equal chords.

🔢 Formula

Important Formulae and Relations

Angle at centre:

\[\small \angle \text{ at centre} = 2 \times \angle \text{ at circumference} \]

Angles in same segment:

\[\small \angle APB = \angle AQB \]

Opposite angles of cyclic quadrilateral:

\[\small \angle A + \angle C = 180^\circ \]

\[\small \angle B + \angle D = 180^\circ \]

Equal chords:

\[\small AB = CD \Rightarrow \angle AOB = \angle COD \]

Equal central angles:

\[\small \angle AOB = \angle COD \Rightarrow AB = CD \]

Perpendicular from centre:

\[\small OM \perp AB \Rightarrow MA = MB \]

ℹ️ Information

Board Examination Focus Areas

Proofs based on congruent triangles are frequently asked.
Cyclic quadrilateral properties are highly important for CBSE board exams.
Converse theorems are commonly asked in competency-based questions.
Theorems related to equal chords and perpendicular distances are important for HOTS questions.
Angle properties of circles are frequently used in case study questions.
Questions involving angle in semicircle are common in objective and MCQ sections.

🧠 Remember

Memory Tricks for Quick Recall

Bigger chord \(\small \Rightarrow \) Bigger arc \( \Rightarrow \) Bigger central angle
Centre angle is always double the circumference angle.
Same segment means same angle.
Diameter at circumference always forms right angle.
Cyclic quadrilateral opposite angles always add to: \[\small 180^\circ \]

❌ Common Mistakes

Common Mistakes to Avoid

Confusing adjacent angles with opposite angles in cyclic quadrilaterals.
Forgetting the \(180^\circ\) relation for opposite angles in cyclic quadrilaterals.
Assuming all quadrilaterals are cyclic without verification.
Mixing up theorems and their converses.
Neglecting to check for cyclicity before applying related theorems.

🏫 NCERT · Class IX · Mathematics

Circles

Chapter 10 — Interactive AI Study Engine. Learn 8 theorems, solve 20+ problems, master every circle concept.

8Theorems

12Formulas

15+Questions

5Interactive

📍

Basic Terms & Definitions

The foundation of circle geometry

✨ Circle

The set of all points in a plane at a fixed distance r (radius) from a fixed point O (centre).

📍 Radius

A line segment joining the centre to any point on the circle. All radii of a circle are equal. OA = OB = OC = r.

〰️ Chord

A line segment joining any two distinct points on the circle. A diameter is the longest chord, passing through the centre.

📏 Diameter

A chord passing through the centre. d = 2r. It bisects the circle into two equal semicircles.

🌙 Arc

A continuous part of the circumference. A semicircle is an arc of exactly half the circle (180°).

🍕 Sector

The region bounded by two radii and the included arc. Area = (θ/360°) × πr².

📐

Key Theorems

8 theorems that define circle geometry — each in its own card

A · Equal Chords & Angles

Theorem 1: Equal Chords subtend Equal Angles at the Centre

Equal chords of a circle subtend equal angles at the centre. Conversely: If angles at the centre are equal, the chords are equal.

B · Perpendicular from Centre

Theorem 2: Perpendicular from Centre Bisects the Chord

The perpendicular from the centre of a circle to a chord bisects the chord. If OM ⊥ AB, then AM = MB.

C · Three Non-Collinear Points

Theorem 3: Unique Circle through Three Points

Only one unique circle passes through any three non-collinear points. This is why the circumcircle of a triangle exists and is unique.

D · Equidistant Chords

Theorem 4: Equal Chords are Equidistant from the Centre

Equal chords of a circle are at the same distance from the centre. Conversely, chords equidistant from the centre are equal in length.

E · Central Angle Theorem

Theorem 5: Angle at Centre = 2 × Angle at Circumference

The angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at any point on the remaining part of the circumference.

Central Angle Theorem∠Central = 2 × ∠Circumference

F · Angles in the Same Segment

Theorem 6: Angles in Same Segment are Equal

Angles subtended by the same arc in the same segment of the circle are equal to one another.

G · Angle in a Semicircle

Theorem 7: Angle in a Semicircle = 90°

The angle subtended by a diameter at the circumference is a right angle (90°). This is a special case of Theorem 5: 180°/2 = 90°.

H · Cyclic Quadrilateral

Theorem 8: Opposite Angles of a Cyclic Quadrilateral are Supplementary

If a quadrilateral is inscribed in a circle, the sum of each pair of opposite angles is 180°. ∠A + ∠C = 180° and ∠B + ∠D = 180°.

🧮

Complete Formula Reference

All key formulas for Circles (Ch. 10)

Geometry of the Circle

Diameterd = 2r

CircumferenceC = 2πr = πd

AreaA = πr²

Arc & Sector

Arc Length (θ in radians)L = r × θ

Arc Length (θ in degrees)L = (θ/360°) × 2πr

Sector Area (θ in radians)S = ½ r²θ

Sector Area (θ in degrees)S = (θ/360°) × πr²

Chord Properties

Chord Length (central θ)c = 2r sin(θ/2)

Distance from centre to chordd = √(r² - (c/2)²)

Radius from chord + distancer² = d² + (c/2)²

Intersection Theorems

Intersecting ChordsAP × PB = CP × PD

Cyclic Quadrilateral∠A + ∠C = 180° and ∠B + ∠D = 180°

Central Angle Theorem∠centre = 2 × ∠circumference

⚙️

Step-by-Step AI Solver

Select a problem type, enter values, get instant guided solutions

👆 Select a problem type above to see what values to enter.

🤖 Solution

📝

Concept-Building Questions

Original problems with full step-by-step solutions — click to reveal

Setup: Three equal-radius circles, pairwise externally tangent. Centroids forms equilateral triangle, each side = 6 cm.

Key insight: Any line through two tangency points is perpendicular to the line joining their centers, at the midpoint.

Distance from center to line: For two tangent circles, the common tangent at the point of contact is the only common point. The "line through two common points" is actually the common tangent shared by two pairs of circles.

For equilateral arrangement: The common tangent to circles at P and Q (two tangency points) cuts the third circle at two points. The distance between farthest intersection points along that line.

Final calculation: With symmetric arrangement and r = 3, the distance between the farthest points on the outer circles along the common tangent = 6 cm.

∴ The required distance = 6 cm

At the centre: ∠AOB = 60°.

For C on major arc: ∠ACB = ½ × ∠AOB = ½ × 60° = 30°.

For D on minor arc: D sees the reflex central angle. Reflex ∠AOB = 360° − 60° = 300°.

Angle at D: ∠ADB = ½ × 300° = 150°.

Verify: ∠ACB + ∠ADB = 30° + 150° = 180°. A, C, B, D form a cyclic quadrilateral — opposite angles are supplementary. ✓

∴ ∠ACB = 30°, ∠ADB = 150°

Central angle: For equilateral triangle, central angle ∠BOC = 360°/3 = 120°.

Sector area: (120°/360°) × π × 7² = (1/3) × 49π = 49π/3 cm².

Chord BC: 2r sin(θ/2) = 2 × 7 × sin(60°) = 14 × √3/2 = 7√3 cm.

Distance: OM = r cos(θ/2) = 7 × cos(60°) = 7 × 0.5 = 3.5 cm.

△BOC area: ½ × 7√3 × 3.5 = 49√3/4 cm².

Segment area = Sector − Triangle: 49π/3 − 49√3/4 = 49(4π − 3√3)/12 cm².

∴ Segment area = 49(4π − 3√3)/12 cm² ≈ 26.87 cm²

Rule: When two chords intersect inside a circle, AP × PB = CP × PD.

Substitute: 6 × 8 = 4 × PD.

Solve: 48 = 4 × PD → PD = 12 cm.

∴ PD = 12 cm (and full chord CD = 4 + 12 = 16 cm)

Given: Trapezium ABCD inscribed in a circle with AB ∥ DC. Need to prove: AD = BC (non-parallel sides equal).

Co-interior angles: AB ∥ DC → ∠DAB + ∠ADC = 180°.

Cyclic property: Opposite angles supplementary: ∠DAB + ∠BCD = 180°.

Therefore: ∠ADC = ∠BCD (both are supplements of ∠DAB).

Equal angles → equal arcs: Equal angles subtend equal arcs. So arcs AD and BC are equal.

Conclusion: Equal arcs have equal chords. Thus AD = BC. ∎

∴ Any cyclic trapezium is always isosceles. ∎

Perpendicular bisects chord: Half-chord = 16/2 = 8 cm.

Right triangle: r² = 6² + 8² = 36 + 64 = 100.

Solve: r = 10 cm.

∴ Radius = 10 cm

🎯

Central Angle Visualizer

Drag the slider to see how central and circumference angles relate

0°

60°

Central ∠ = 60° → Circumference ∠ = 30°

⚖️

Chord Distance Calculator

Enter chord length and distance from centre to find the radius

🤖 Result

♻️

Cyclic Quad Angle Slider

Change ∠A and see ∠C update (A + C = 180°)

🃏

Theorem Flash Cards

Tap a card to flip and reveal the theorem

📏Perpendicular from Centre

A perpendicular from the centre to a chord bisects the chord. OM ⊥ AB → AM = MB.

📐Central Angle Theorem

∠ at centre = 2 × ∠ at circumference. The same arc subtends angle double at the centre.

♻️Cyclic Quadrilateral

Opposite angles of a cyclic quadrilateral are supplementary. ∠A + ∠C = 180°.

🔲Angle in Semicircle

The angle subtended by a diameter at any point on the circumference is always 90°.

⚖️Equal Chords

Equal chords are equidistant from the centre, and vice versa.

✂️Intersecting Chords

When two chords intersect inside a circle: AP × PB = CP × PD.

🌙Same Segment

Angles in the same segment of a circle are always equal.

📍Three Points

Exactly one circle passes through any three non-collinear points.

🏆

Quick Concept Check

8 concept-testing questions with instant feedback

✏️

Fill in the Blanks

Type your answer and see if it's correct

💡

Tips & Tricks

Smart strategies to solve circle problems faster

🎯Always draw the diagram first. Label the centre (O), points (A, B, C...), and mark what's given. A good diagram solves 50% of the problem.
📐Central angle theorem: double at centre, half at circumference. This is the most used theorem. Memorise it both ways.
⚖️When you see "perpendicular from centre to chord," immediately write: bisects chord + right triangle + Pythagoras.
♻️For cyclic quadrilaterals: opposite angles are supplementary. Always check: ∠A + ∠C = 180° and ∠B + ∠D = 180°.
🔲Angle in a semicircle = 90°. When you see a diameter, mark the right angle BEFORE doing anything else. It's almost always the key step.
🔗Intersecting chords: AP × PB = CP × PD. This is just the power of a point theorem. If you know three of the four segments, you can always find the fourth.
🧩Equal chords → equal arcs → equal central angles → equal inscribed angles. Chain these equalities like a shortcut.

⚠️

Common Mistakes to Avoid

Every student makes these — you don't have to

❌Forgetting to halve. Angle at the centre = 2 × angle at circumference, not equal. Students often write 80° for both — always divide by 2.
❌Using reflex angle incorrectly. Points on the major arc use the minor arc's central angle; points on the minor arc use the reflex angle (360° − minor). Check which arc your point belongs to.
❌Assuming any bisector is from the centre. Only the perpendicular bisector FROM THE CENTRE has special properties. Random perpendiculars don't necessarily bisect the chord.
❌Confusing cyclic quad rules. Opposite angles sum to 180°, not adjacent. Adjacent angles add to whatever they add to — only opposite ones are related.
❌Mixing chord and arc length. Chord = straight line between two points. Arc = curved path. Different formulas. Don't substitute one for the other.
❌Forgetting diameter rules. "Angle in a semicircle = 90°" only works for a DIAMETER. If the chord isn't a diameter, the angle is NOT 90°.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025

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NCERT Class 9 Maths Chapter 9 Circles Notes

NCERT Class 9 Maths Chapter 9 Circles Notes — Complete Notes & Solutions · academia-aeternum.com

Chapter 9, “Circles,” introduces students to one of the most fundamental and visually interesting shapes in geometry. A circle is more than just a round figure — it is a geometric form defined by a set of points that remain at the same distance from a fixed point called the centre. This chapter helps students explore important terms such as radius, diameter, arc, chord, sector, segment, and central angle. Using logical reasoning and simple diagrams, the chapter builds a clear understanding of…

🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE

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