NCERT Solutions for Class 9 Maths Chapter 9: Circles — Free PDF

Chapter 9 is a substantial geometry chapter that covers the properties of circles, chords, arcs, and cyclic quadrilaterals. It builds heavily on the triangle congruence skills you developed in Chapter 7 — most proofs in this chapter involve proving two triangles congruent and then using CPCT.

The chapter introduces several powerful theorems about circles that form the foundation of all circle geometry you will encounter in Class 10 (tangent properties) and beyond. These theorems relate chords to their distances from the centre, arcs to their subtended angles, and establish the remarkable angle properties of cyclic quadrilaterals.

The chapter has six exercises with a strong focus on theorem-based proofs. Exercise 9.1 covers the angle subtended by a chord at the centre. Exercise 9.2 deals with the perpendicular from the centre to a chord. Exercise 9.3 covers the relationship between central angles and inscribed angles. Exercise 9.4 focuses on cyclic quadrilaterals. Exercises 9.5 and 9.6 contain additional problems combining multiple theorems. This chapter typically carries 6-8 marks in CBSE exams.

Circle: The set of all points in a plane that are at a fixed distance (radius) from a fixed point (centre).

Chord: A line segment joining any two points on a circle. The diameter is the longest chord, passing through the centre.

Arc: A continuous piece of a circle. A chord divides a circle into two arcs — the major arc (the larger one) and the minor arc (the smaller one).

Segment: The region between a chord and the arc it cuts off. The major segment is the larger region, and the minor segment is the smaller one.

Sector: The region between two radii and the arc they intercept (like a slice of pizza).

Central Angle: An angle subtended by an arc at the centre of the circle.

Inscribed Angle: An angle subtended by an arc at any point on the remaining part of the circle.

Cyclic Quadrilateral: A quadrilateral whose all four vertices lie on a single circle. Its key property is that the sum of each pair of opposite angles is $180^\circ$.

Concyclic Points: Points that lie on the same circle.

Theorem: Equal chords of a circle subtend equal angles at the centre.

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Problem 1: If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of the other chord.

Solution:
Let chords $AB$ and $CD$ intersect at $P$ inside the circle. Given $AB = CD$.

Draw $OM \perp AB$ and $ON \perp CD$ from centre $O$.

Since $AB = CD$, equal chords are equidistant from the centre:

Also, $M$ is the midpoint of $AB$ and $N$ is the midpoint of $CD$ (perpendicular from centre bisects chord).

In $\triangle OMP$ and $\triangle ONP$:
- $OM = ON$ (proved)
- $OP = OP$ (common)
- $\angle OMP = \angle ONP = 90^\circ$

By RHS: $\triangle OMP \cong \triangle ONP$

By CPCT: $MP = NP$

Since $AM = \dfrac{1}{2}AB = \dfrac{1}{2}CD = CN$:

So $AP = CP$ and $BP = DP$. $\blacksquare$

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Problem 2: Two chords $AB$ and $CD$ of lengths $5$ cm and $11$ cm respectively of a circle are parallel to each other and are on opposite sides of the centre. If the distance between them is $6$ cm, find the radius of the circle.

Solution:
Draw $OM \perp AB$ and $ON \perp CD$ from centre $O$.

$AM = \dfrac{5}{2} = 2.5$ cm and $CN = \dfrac{11}{2} = 5.5$ cm.

Since the chords are on opposite sides: $MN = OM + ON = 6$ cm.

Let $OM = x$, then $ON = 6 - x$.

In $\triangle OMA$: $r^2 = OM^2 + AM^2 = x^2 + 6.25$

In $\triangle ONC$: $r^2 = ON^2 + CN^2 = (6-x)^2 + 30.25$

Equating: $x^2 + 6.25 = (6-x)^2 + 30.25$

$x^2 + 6.25 = 36 - 12x + x^2 + 30.25$

$6.25 = 66.25 - 12x$

$12x = 60 \implies x = 5$

$r^2 = 25 + 6.25 = 31.25$

$r = \sqrt{31.25} = \dfrac{5\sqrt{5}}{2} \approx 5.59$ cm.

Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.

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Problem 1: Two circles of radii $5$ cm and $3$ cm intersect at two points and the distance between their centres is $4$ cm. Find the length of the common chord.

Solution:
Let the centres be $O$ and $O'$ with radii $r_1 = 5$ cm and $r_2 = 3$ cm. Let $OO' = 4$ cm.

Let the common chord be $AB$ and let $M$ be the midpoint of $AB$ (since $OO' \perp AB$ and bisects it).

Let $OM = x$, so $O'M = 4 - x$.

In $\triangle OMA$: $OA^2 = OM^2 + MA^2$

In $\triangle O'MA$: $O'A^2 = O'M^2 + MA^2$

Subtract (ii) from (i):

So $OM = 4 = OO'$, meaning $O'$ lies on the chord.

From (i): $MA^2 = 25 - 16 = 9$, so $MA = 3$ cm.

$AB = 2 \times MA = 6$ cm.

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Problem 2: A chord of length $16$ cm is drawn in a circle of radius $10$ cm. Find the distance of the chord from the centre.

Solution:
Let $M$ be the midpoint of the chord. Then $AM = 8$ cm.

In $\triangle OMA$ (right-angled at $M$):

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Problem 3: Two chords of a circle are each $10$ cm long. If the radius is $13$ cm, are they equidistant from the centre? Find the distance.

Solution:
Since the chords are equal in length, by the theorem "equal chords are equidistant from the centre", they are equidistant.

For each chord: half-length $= 5$ cm.

Distance $= \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

Theorem (Central Angle Theorem): 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.

If $\angle AOB$ is the central angle and $\angle ACB$ is the inscribed angle subtending the same arc:

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Problem 1: In the figure, $\angle PQR = 100^\circ$, where $P$, $Q$, $R$ are points on a circle with centre $O$. Find $\angle OPR$.

Solution:
Reflex $\angle POR = 2 \times \angle PQR = 2 \times 100^\circ = 200^\circ$

$\therefore \angle POR = 360^\circ - 200^\circ = 160^\circ$

In $\triangle OPR$: $OP = OR$ (radii)

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Problem 2: $A$, $B$, $C$, $D$ are four points on a circle. $AC$ and $BD$ intersect at $E$ such that $\angle BEC = 130^\circ$ and $\angle ECD = 20^\circ$. Find $\angle BAC$.

Solution:
In $\triangle DEC$:
$\angle DEC = 180^\circ - \angle BEC = 180^\circ - 130^\circ = 50^\circ$ (vertically opposite to $\angle AEB$, and $\angle DEC = \angle AEB = 50^\circ$... Actually $\angle DEC$ and $\angle BEC$ are supplementary since they form a linear pair on line $AC$. Wait — $\angle DEC$ is vertically opposite to $\angle AEB$. Since $\angle BEC = 130^\circ$, $\angle AEB = 50^\circ$ (linear pair), and $\angle DEC = 50^\circ$ (vertically opposite to $\angle AEB$).

In $\triangle DEC$: $\angle DEC + \angle ECD + \angle EDC = 180^\circ$

Now $\angle BDC = \angle EDC = 110^\circ$.

$\angle BAC$ and $\angle BDC$ are angles subtended by the same chord $BC$ in the same segment.

$\therefore \angle BAC = \angle BDC = 110^\circ$.

Wait — for angles in the same segment, they must be on the same side. If $A$ and $D$ are on the same arc (same side of chord $BC$), then $\angle BAC = \angle BDC$. Since $ABCD$ are concyclic and the angles subtend the same arc, $\angle BAC = \angle BDC = 110^\circ$.

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Problem 3: If $AB$ is a diameter and $C$ is any point on the circle, prove that $\angle ACB = 90^\circ$.

Solution:
Since $AB$ is a diameter, the arc $AB$ (going the long way round) is a semicircle.

The central angle subtended by this arc $= 180^\circ$ (a straight line through the centre).

By the central angle theorem: $\angle ACB = \dfrac{1}{2} \times 180^\circ = 90^\circ$. $\blacksquare$

This is the famous angle in a semicircle theorem and is one of the most frequently tested results.

Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$.

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Problem 1: If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:
Let $ABCD$ be a trapezium with $AB \parallel DC$ and $AD = BC$.

Draw $AM \perp DC$ and $BN \perp DC$.

In $\triangle AMD$ and $\triangle BNC$:
- $AD = BC$ (given)
- $AM = BN$ (perpendicular distance between parallel lines is constant)
- $\angle AMD = \angle BNC = 90^\circ$

By RHS: $\triangle AMD \cong \triangle BNC$

By CPCT: $\angle ADC = \angle BCD$

Now $\angle DAB + \angle ADC = 180^\circ$ (co-interior angles, $AB \parallel DC$)

$\therefore \angle DAB + \angle BCD = 180^\circ$

Since the sum of opposite angles is $180^\circ$, $ABCD$ is a cyclic quadrilateral. $\blacksquare$

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Problem 2: $ABCD$ is a cyclic quadrilateral. If $\angle A = 3\angle C$, find $\angle A$.

Solution:

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Problem 3: In a cyclic quadrilateral $ABCD$, $\angle A = (2x + 4)^\circ$, $\angle B = (y + 3)^\circ$, $\angle C = (2y + 10)^\circ$, $\angle D = (4x - 5)^\circ$. Find all four angles.

Solution:
Opposite angles sum to $180^\circ$:

$\angle A + \angle C = 180^\circ$: $(2x + 4) + (2y + 10) = 180 \implies 2x + 2y = 166 \implies x + y = 83 \quad \cdots (1)$

$\angle B + \angle D = 180^\circ$: $(y + 3) + (4x - 5) = 180 \implies 4x + y = 182 \quad \cdots (2)$

Subtract (1) from (2): $3x = 99 \implies x = 33$.

From (1): $y = 83 - 33 = 50$.

$\angle A = 2(33) + 4 = 70^\circ$, $\angle B = 50 + 3 = 53^\circ$, $\angle C = 2(50) + 10 = 110^\circ$, $\angle D = 4(33) - 5 = 127^\circ$.

Verification: $70 + 110 = 180$ and $53 + 127 = 180$. $\checkmark$

Example 1: $O$ is the centre of a circle. $\angle BOC = 60^\circ$ and $\angle AOB = 100^\circ$. Find $\angle ABC$.

Solution:
$\angle AOC = 360^\circ - 100^\circ - 60^\circ = 200^\circ$ (angles around $O$).

The inscribed angle $\angle ABC$ subtends arc $AC$ (the arc not containing $B$).

The central angle subtending the same arc is $\angle AOC = 200^\circ$.

By the central angle theorem: $\angle ABC = \dfrac{1}{2} \times 200^\circ = 100^\circ$.

Wait — the reflex angle at the centre is $200^\circ$. The inscribed angle on the major arc would be $\dfrac{200}{2} = 100^\circ$. But $B$ is on the minor arc of $AC$, so the arc $AC$ that does not contain $B$ has central angle $200^\circ$. The inscribed angle from $B$ uses the reflex: $\angle ABC = \dfrac{1}{2}(\text{reflex } \angle AOC)$. Actually, the non-reflex $\angle AOC = 360 - 200 = 160^\circ$. Hmm, let me reconsider.

$\angle AOB = 100^\circ$ and $\angle BOC = 60^\circ$. So $\angle AOC = 360 - 100 - 60 = 200^\circ$ is the reflex angle.

The non-reflex $\angle AOC = 100 + 60 = 160^\circ$... no. The three arcs around the circle are: arc $AB$ (central angle $100^\circ$), arc $BC$ (central angle $60^\circ$), arc $CA$ (central angle $200^\circ$). The inscribed angle $\angle ABC$ subtends arc $ADC$ (the arc not containing $B$), which has central angle $200^\circ$.

So $\angle ABC = \dfrac{1}{2} \times 200^\circ = 100^\circ$.

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Example 2: Prove that the quadrilateral formed by the internal bisectors of angles of a cyclic quadrilateral is also cyclic.

Solution:
Let $ABCD$ be a cyclic quadrilateral. Let the internal bisectors of $\angle A$, $\angle B$, $\angle C$, $\angle D$ form quadrilateral $PQRS$.

Since $ABCD$ is cyclic: $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$.

So $\dfrac{\angle A}{2} + \dfrac{\angle C}{2} = 90^\circ$ and $\dfrac{\angle B}{2} + \dfrac{\angle D}{2} = 90^\circ$.

In $\triangle APB$: $\angle P = 180^\circ - \dfrac{\angle A}{2} - \dfrac{\angle B}{2}$.

In $\triangle CRD$: $\angle R = 180^\circ - \dfrac{\angle C}{2} - \dfrac{\angle D}{2}$.

$\angle P + \angle R = 360^\circ - \dfrac{\angle A + \angle C}{2} - \dfrac{\angle B + \angle D}{2} = 360^\circ - 90^\circ - 90^\circ = 180^\circ$.

Since opposite angles of $PQRS$ sum to $180^\circ$, it is cyclic. $\blacksquare$

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Example 3: $AB$ is a chord of a circle with centre $O$. $AB = 10$ cm and the radius is $13$ cm. Find the distance of the chord from the centre, and the angle $\angle AOB$.

Solution:
Let $M$ be the midpoint of $AB$. Then $AM = 5$ cm.

$OM = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

In $\triangle OMA$: $\sin(\angle AOM) = \dfrac{AM}{OA} = \dfrac{5}{13}$.

$\angle AOM = \sin^{-1}\left(\dfrac{5}{13}\right) \approx 22.6^\circ$.

$\angle AOB = 2 \times \angle AOM \approx 45.2^\circ$.

Mistake 1: Applying the central angle theorem incorrectly.
The theorem states: central angle $= 2 \times$ inscribed angle, but only when both angles subtend the same arc. Make sure you identify the correct arc.

Mistake 2: Confusing major and minor arcs.
A chord creates two arcs. The inscribed angle depends on which arc the point lies on. Points on the major arc subtend smaller angles than points on the minor arc for the same chord.

**Mistake 3: Forgetting that the angle in a semicircle is $90^\circ$.**
If $AB$ is a diameter and $C$ is on the circle, then $\angle ACB = 90^\circ$. This is one of the most frequently tested facts and is often missed in multi-step problems.

Mistake 4: Assuming a quadrilateral is cyclic without proof.
You must either prove that opposite angles sum to $180^\circ$ or show that all four vertices lie on a common circle. Do not assume cyclic property without justification.

Mistake 5: Not drawing radii to create isosceles triangles.
In circle problems, drawing radii to relevant points creates isosceles triangles (all radii are equal), which is the key technique for finding unknown angles.

Q1. In a circle with centre $O$, $\angle AOB = 80^\circ$. Find $\angle ACB$ where $C$ is a point on the major arc.

Answer: $\angle ACB = \dfrac{1}{2} \times 80^\circ = 40^\circ$ (central angle theorem).

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Q2. $ABCD$ is a cyclic quadrilateral with $\angle A = 110^\circ$ and $\angle B = 70^\circ$. Find $\angle C$ and $\angle D$.

Answer: $\angle C = 180^\circ - 110^\circ = 70^\circ$ and $\angle D = 180^\circ - 70^\circ = 110^\circ$.

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Q3. A chord of length $24$ cm is at a distance of $5$ cm from the centre. Find the radius.

Answer: Half-chord $= 12$ cm. $r = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ cm.

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Q4. $AB$ is a diameter. $\angle BAC = 35^\circ$. Find $\angle ABC$.

Answer: Since $AB$ is a diameter, $\angle ACB = 90^\circ$. In $\triangle ABC$: $\angle ABC = 180^\circ - 90^\circ - 35^\circ = 55^\circ$.

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Q5. Two chords $AB$ and $CD$ are equal. If the distance of $AB$ from the centre is $4$ cm and the radius is $5$ cm, find the length of each chord.

Answer: Half-chord $= \sqrt{5^2 - 4^2} = \sqrt{9} = 3$ cm. Each chord $= 6$ cm.

• The perpendicular from the centre to a chord bisects it — this is the most-used construction in circle problems.
  - Equal chords are equidistant from the centre, and vice versa.
  - Central angle $= 2 \times$ inscribed angle (on the same arc) is the most powerful theorem.
  - Angle in a semicircle is always $90^\circ$.
  - Angles in the same segment (subtending the same chord from the same side) are equal.
  - Opposite angles of a cyclic quadrilateral sum to $180^\circ$.
  - Most proofs involve drawing radii to create isosceles triangles, then using congruence (from Chapter 7) and CPCT.
  - These theorems form the foundation for Class 10 circles (tangent properties).

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