NCERT Class 9 Circles Exercise 9.1 Solutions

📘 Concept & Theory Theory & Concepts ›

1. Congruent Circles

Two circles are said to be congruent if and only if they have the same radius. This means their corresponding points can be made to coincide exactly when one circle is placed over the other.

If two circles have radii \(\small r_1 \) and \(\small r_2 \), they are congruent if and only if \(\small r_1 = r_2 \).

2. Equal Chords

Two chords of a circle (or congruent circles) are said to be equal or congruent if they have the same length. In the problem, we are given that \(\small PQ = P'Q' \).

3. Angle Subtended at Centre

When we draw radii from the centre of a circle to the endpoints of a chord, an angle is formed at the centre. This angle is called the angle subtended by the chord at the centre.

In the given problem:

Chord \(\small PQ \) subtends \(\small \angle POQ \) at centre \(\small O \)
Chord \(\small P'Q' \) subtends \(\small \angle P'O'Q' \) at centre \(\small O' \)

4. SSS Congruence Criterion

If three sides of one triangle are respectively equal to the three sides of another triangle, then the two triangles are congruent (SSS: Side-Side-Side).

For congruence of \(\small \triangle OPQ \) and \(\small \triangle O'P'Q' \), we need to establish: \[\small OP = O'P', \quad OQ = O'Q', \quad PQ = P'Q' \]

5. CPCT (Corresponding Parts of Congruent Triangles)

When two triangles are proven congruent by any criterion (SSS, SAS, ASA, AAS, or RHS), we can state that all corresponding parts are equal. This is CPCT — Corresponding Parts of Congruent Triangles.

Specifically, if \(\small \triangle ABC \cong \triangle DEF \), then: \[\small \begin{aligned} \angle A &= \angle D, \\ \angle B &= \angle E, \\ \angle C &= \angle F \end{aligned}\] \[\small \begin{aligned} AB &= DE, \\ BC &= EF, \\ CA &= FD \end{aligned}\]

🗺️ Solution Roadmap Step-by-step Plan ›

Identify Given Information

Congruent circles (equal radii \(\small r \)) and equal chords \(\small PQ = P'Q' \)
Form Triangles

Draw triangles \(\small \triangle OPQ \) and \(\small \triangle O'P'Q' \) using radii to chord endpoints
Establish SSS Conditions

Show: \(\small OP = O'P' \), \(\small OQ = O'Q' \), \(\small PQ = P'Q' \)
Apply SSS Congruence

Conclude \(\small \triangle OPQ \cong \triangle O'P'Q' \)
Apply CPCT

Corresponding angles are equal: \(\small \angle POQ = \angle P'O'Q' \)

📊 Graph / Figure Graph / Figure ›

Two congruent circles with equal chords PQ and P'Q' subtending angles ∠POQ and ∠P'O'Q' at their respective centres

📐 Proof Proof ›

📌 Given

Two congruent circles with centres \(\small O \) and \(\small O' \)
Radius of each circle is \(\small r \) (since circles are congruent, radii are equal)
Two chords \(\small PQ \) and \(\small P'Q' \) such that their lengths are equal: \[\small PQ = P'Q' \]

🎯 To Prove

The angles subtended by chords \(\small PQ \) and \(\small P'Q' \) at the centres of their respective circles are equal, i.e.: \[\small \angle POQ = \angle P'O'Q' \]

🔧 Construction

Join the centre of each circle to the endpoints of the chords to form two triangles:

In Circle 1: Join \(\small OP \), \(\small OQ \) (radii)
In Circle 2: Join \(\small O'P' \), \(\small O'Q' \) (radii)

This gives us two triangles: \(\small \triangle OPQ \) and \(\small \triangle O'P'Q' \)

✍️ Proof

Step-by-step Proof · 14 steps · 6 parts

Step 1: Identify the triangles formed
In Circle 1, we have triangle \(\small \triangle OPQ \) with sides \(\small OP \), \(\small OQ \), and \(\small PQ \).
In Circle 2, we have triangle \(\small \triangle O'P'Q' \) with sides \(\small O'P' \), \(\small O'Q' \), and \(\small P'Q' \).
We need to prove that these two triangles are congruent to establish that \(\small \angle POQ = \angle P'O'Q' \).
Step 2: Establish first side equality
Since both circles are congruent (having equal radii), and \(\small OP \) and \(\small O'P' \) are radii of their respective circles: \[\small OP = O'P' = r \]
This gives us the first side equality: \(\small OP = O'P' \)
Step 3: Establish second side equality
Similarly, \(\small OQ \) and \(\small O'Q' \) are also radii of their respective congruent circles: \[\small OQ = O'Q' = r \]
This gives us the second side equality: \(\small OQ = O'Q' \)
Step 4: Establish third side equality
The chords are given to be equal: \[\small PQ = P'Q' \]
This gives us the third side equality: \(\small PQ = P'Q' \)
Step 5: Apply SSS Congruence Criterion
\[\small OP = O'P' \quad \text{(radii of congruent circles)} \] \[\small OQ = O'Q' \quad \text{(radii of congruent circles)} \] \[\small PQ = P'Q' \quad \text{(given: equal chords)} \]
Now we have established all three pairs of equal sides:
Therefore:
\[\small \triangle OPQ \cong \triangle O'P'Q' \]
SSS (Side-Side-Side) Congruence Criterion
Step 6: Apply CPCT (Corresponding Parts of Congruent Triangles)
Since \(\small \triangle OPQ \cong \triangle O'P'Q' \), all corresponding parts of these congruent triangles are equal.
The angle at vertex O (between sides OP and OQ) corresponds to the angle at vertex O' (between sides O'P' and O'Q'). \[\small \angle POQ = \angle P'O'Q' \]

Q.E.D.

📝 Key Takeaways ›

Congruent circles have equal radii
SSS Rule: All three sides equal → Triangles are congruent
CPCT: Congruent triangles have all corresponding parts equal
The theorem works both ways (direct and converse)
This theorem is fundamental to understanding the relationship between chord length and the angle subtended at the centre

🎯 Exam Significance Exam Significance ›

Frequently asked as a proof-based question in CBSE Class 9 final exams
Directly from NCERT — Exercise 9.1, Question 1, a fundamental theorem proof
2-3 marks guaranteed if asked as a separate proof question
Often combined with other circle theorems to create 4-6 mark proof chains

📘 Concept & Theory Theory & Concepts ›

1. Converse of a Theorem

This question is the converse of Q1 (Exercise 9.1). Q1 established that:
"Equal chords subtend equal angles at the centre."
Q2 asks us to prove the reverse direction:
"Equal angles at the centre imply equal chords."

2. SAS Congruence Criterion

SAS (Side-Angle-Side): If two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the triangles are congruent.
Conditions needed:

Side 1: \(\small AB = PQ \) ✔
Included Angle: \(\small \angle BAC = \angle QPR \) ✔
Side 2: \(\small AC = PR \) ✔

3. CPCT (Corresponding Parts of Congruent Triangles)

After establishing triangle congruence via SAS, all corresponding parts are equal. Since \(\small \triangle ABC \cong \triangle PQR \), we get: \[\small BC = QR \]

4. Logical Structure

Together, Q1 and Q2 prove a biconditional relationship:
Equal chords \(\small \iff \) Equal angles at centre
This is a fundamental property used extensively in circle geometry.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify Given Information: Congruent circles (equal radii), chords BC and QR subtending equal angles at centres.
Form Triangles: Draw triangles \(\small \triangle ABC \) and \(\small \triangle PQR \) using radii to chord endpoints.
Establish Side Equality: Show \(\small AB = PQ \) and \(\small AC = PR \) (radii of congruent circles).
Establish Angle Equality: Confirm \(\small \angle BAC = \angle QPR \) (given condition).
Apply SAS Congruence: Since two sides and included angle are equal, conclude \(\small \triangle ABC \cong \triangle PQR \).
Apply CPCT: Corresponding sides of congruent triangles are equal, so \(\small BC = QR \)
.

📊 Graph / Figure Graph / Figure ›

Figure: Two congruent circles with chords BC and QR subtending equal angles ∠BAC and ∠QPR at their respective centres

📐 Proof Proof ›

📌 Given

Two congruent circles with centres \(\small A \) and \(\small P \)
Radius of each circle is \(\small r \) (since circles are congruent, radii are equal)
Two chords \(\small BC \) and \(\small QR \) such that the angles subtended by these chords at their respective centres are equal: \[\small \angle BAC = \angle QPR \]

🎯 To Prove

The chords \(\small BC \) and \(\small QR \) are equal, i.e.: \[\small BC = QR \]

✍️ Proof

Step-by-step Proof · 23 steps · 6 parts

Step 1: Identify the triangles formed
In Circle 1 with centre \(\small A \), joining radii to chord endpoints gives triangle \(\small \triangle ABC \) with sides \(\small AB \), \(\small AC \), and chord \(\small BC \).
>In Circle 2 with centre \(\small P \), joining radii to chord endpoints gives triangle \(\small \triangle PQR \) with sides \(\small PQ \), \(\small PR \), and chord \(\small QR \).
To prove \(\small BC = QR \), we will first prove \(\small \triangle ABC \cong \triangle PQR \) using SAS criterion.
Step 2: Establish first side equality (AB = PQ)
Since both circles are congruent, they have equal radii.
In Circle 1, \(\small AB \) is a radius joining centre \(\small A \) to point \(\small B \) on the circle. \[\small AB = r \]
In Circle 2, \(\small PQ \) is a radius joining centre \(\small P \) to point \(\small Q \) on the circle. \[\small PQ = r \]
Therefore:\[\small AB = PQ \]
Step 3: Establish second side equality (AC = PR)
Similarly, \(\small AC \) is a radius in Circle 1: \[\small AC = r \]
And \(\small PR \) is a radius in Circle 2: \[\small PR = r \]
Therefore: \[\small AC = PR \]
Step 4: Verify the included angle equality (\(\small \angle BAC = \angle QPR \))
The given condition states that chords \(\small BC \) and \(\small QR \) subtend equal angles at their respective centres.
In Circle 1, chord \(\small BC \) subtends angle \(\small \angle BAC \) at centre \(\small A \). This is the angle between radii \(\small AB \) and \(\small AC \), hence it is the included angle.
In Circle 2, chord \(\small QR \) subtends angle \(\small \angle QPR \) at centre \(\small P \). This is the angle between radii \(\small PQ \) and \(\small PR \), hence it is also the included angle.
By the given condition: \[\small \angle BAC = \angle QPR \]
Step 5: Apply SAS Congruence Criterion
\[\small \boxed{AB = PQ} \quad \text{(radii of congruent circles)} \]
\[\small \boxed{\angle BAC = \angle QPR} \quad \text{(given: equal angles subtended)} \]
\[\small \boxed{AC = PR} \quad \text{(radii of congruent circles)} \]
Two sides and the included angle of \(\small \triangle ABC \) are respectively equal to two sides and the included angle of \(\small \triangle PQR \).
Therefore, by the SAS (Side-Angle-Side) Congruence Criterion: \[\small \triangle ABC \cong \triangle PQR \]
Step 6: Apply CPCT (Corresponding Parts of Congruent Triangles)
Since \(\small \triangle ABC \cong \triangle PQR \), all corresponding parts of the two congruent triangles are equal.
In \(\small \triangle ABC \), the side opposite to \(\small \angle BAC \) is \(\small BC \).
In \(\small \triangle PQR \), the side opposite to \(\small \angle QPR \) is \(\small QR \).
By CPCT, the corresponding sides opposite to the equal angles are also equal: \[\small BC = QR \]