Class 9Maths · GeometryFull chapter

Circles

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Circles and Related TermsPerpendicular from the Centre to a ChordEqual Chords and Equidistance from the CentreAngle Subtended by an Arc and the Angle in a SemicircleCyclic Quadrilaterals

Perpendicular from the Centre to a Chord

Quick answer The perpendicular drawn from the centre of a circle to a chord always bisects the chord, and this fact combines with the Pythagoras theorem to relate the radius, the chord length and its distance from the centre.

Every chord of a circle has a definite relationship with the centre through the perpendicular drawn from the centre to it.

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

Converse: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

This gives a very useful right triangle: if O is the centre, AB is a chord, and OM ⊥ AB (M on AB), then AM = MB, and triangle OMA is right-angled at M, so OA² = OM² + AM² (Pythagoras theorem). This single relation connects the radius, the distance of the chord from the centre, and half the length of the chord, and is used repeatedly to find any one of these three quantities when the other two are known.

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

Since the perpendicular from the centre bisects the chord, AM = 16/2 = 8 cm.

In right triangle OMA: OA² = OM² + AM²

10² = OM² + 8²

100 = OM² + 64

OM² = 36, so OM = 6 cm.

The chord is 6 cm from the centre.

Worked Example: Find the length of a chord that is 8 cm from the centre of a circle of radius 17 cm.

Let M be the midpoint of the chord AB, so OM = 8 cm and OA = 17 cm.

OA² = OM² + AM²

17² = 8² + AM²

289 = 64 + AM²

AM² = 225, so AM = 15 cm.

Since OM bisects AB, AB = 2 × AM = 30 cm.

Right triangle from centre to chord OA² = OM² + AM² O = centre, M = midpoint of chord AB, OM ⊥ AB
Half chord length AM = AB/2
Remember
  • The perpendicular from the centre of a circle to a chord bisects the chord.
  • Conversely, the line joining the centre to the midpoint of a chord is perpendicular to the chord.
  • If M is the midpoint of chord AB and O is the centre, triangle OMA is right-angled at M.
  • OA² = OM² + AM² connects the radius, the distance from the centre, and half the chord — use it to find any one quantity from the other two.
  • This theorem is the key tool for solving numerical problems involving chords, radii and distances from the centre.

Equal Chords and Equidistance from the Centre

Quick answer Chords of equal length in a circle (or congruent circles) lie at equal distances from the centre, and this relationship works both ways.

Theorem: Equal chords of a circle (or of congruent circles) are equidistant from the centre.

Converse: Chords of a circle (or of congruent circles) that are equidistant from the centre are equal in length.

Here "distance from the centre" means the length of the perpendicular drawn from the centre to the chord.

Worked Example: Two chords AB and CD of a circle with centre O are equal in length. If AB = 24 cm and the radius of the circle is 13 cm, find the distance of AB from the centre, and state the distance of CD from the centre.

Half of AB = 24/2 = 12 cm.

Using OA² = OM² + AM²: 13² = OM² + 12²

169 = OM² + 144

OM² = 25, so OM = 5 cm.

So AB is 5 cm from the centre. Since AB = CD (equal chords), CD is also 5 cm from the centre, by the theorem that equal chords are equidistant from the centre.

Distance-chord relation r² = d² + (l/2)² r = radius, d = distance from centre, l = chord length
Remember
  • Distance of a chord from the centre = length of the perpendicular from the centre to the chord.
  • Equal chords of a circle are equidistant from the centre.
  • Chords equidistant from the centre are equal in length (converse).
  • This applies to chords within one circle, or corresponding chords of congruent circles.
  • Combine with Pythagoras (r² = d² + (l/2)²) to find unknown distances or chord lengths.

Angle Subtended by an Arc and the Angle in a Semicircle

Quick answer The angle an arc subtends at the centre is double the angle it subtends at any point on the remaining part of the circle, and every angle in a semicircle is a right angle.

An arc of a circle subtends an angle at the centre as well as at any point on the remaining part of the circle.

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.

An important consequence follows for angles in the same segment:

Theorem: Angles in the same segment of a circle are equal.

Theorem: The angle subtended by a semicircle (a diameter) at any point on the circle is a right angle, i.e. the angle in a semicircle is 90°.

Worked Example: An arc AB of a circle subtends an angle of 70° at the centre O. Find the angle subtended by the same arc at a point P on the major arc.

By the theorem, angle at centre = 2 × angle at point on remaining circle.

70° = 2 × angle APB

angle APB = 35°.

Worked Example: AB is a diameter of a circle and C is a point on the circle. Find angle ACB.

Since AB is a diameter, arc ACB is a semicircle and it subtends an angle of 180° at the centre. So the angle subtended at point C on the remaining part of the circle = 180°/2 = 90°. Hence angle ACB = 90°.

Central angle vs circle angle ∠(centre) = 2 × ∠(remaining part of circle)
Angle in a semicircle ∠ACB = 90° AB is a diameter, C lies on the circle
Remember
  • The angle subtended by an arc at the centre is double the angle it subtends at any point on the remaining part of the circle.
  • Angles in the same segment of a circle (subtended by the same arc) are equal.
  • The angle in a semicircle is always a right angle (90°).
  • A diameter subtends a 180° angle at the centre, giving 90° at the circle.
  • These theorems are used together to find unknown angles in circle-based figures.

Cyclic Quadrilaterals

Quick answer A cyclic quadrilateral has all four vertices on a circle, and its opposite angles always add up to 180°.

A quadrilateral is called a cyclic quadrilateral if all four of its vertices lie on a circle.

Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

Converse: If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

Worked Example: ABCD is a cyclic quadrilateral in which angle A = 100° and angle B = 80°. Find angle C and angle D.

Angle A and angle C are opposite angles, so angle A + angle C = 180°.

100° + angle C = 180°, so angle C = 80°.

Angle B and angle D are opposite angles, so angle B + angle D = 180°.

80° + angle D = 180°, so angle D = 100°.

Worked Example: In a cyclic quadrilateral PQRS, angle P = (3x)° and angle R = (2x + 10)°. Find x and angle P.

Since P and R are opposite angles: angle P + angle R = 180°.

3x + 2x + 10 = 180

5x = 170, so x = 34

angle P = 3 × 34 = 102°.

Opposite angles of cyclic quadrilateral ∠A + ∠C = 180°, ∠B + ∠D = 180°
Exterior angle property exterior angle = interior opposite angle
Remember
  • A cyclic quadrilateral has all four vertices lying on a single circle.
  • The sum of each pair of opposite angles of a cyclic quadrilateral is 180°.
  • If opposite angles of a quadrilateral add up to 180°, the quadrilateral must be cyclic.
  • An exterior angle of a cyclic quadrilateral equals the interior opposite angle.
  • These properties are widely used with the arc-angle theorems to solve circle problems.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

d = 2r
Diameter
C = 2πr
Circumference
A = πr²
Area of circle
OA² = OM² + AM²
Right triangle from centre to chord
AM = AB/2
Half chord length
r² = d² + (l/2)²
Distance-chord relation
∠(centre) = 2 × ∠(remaining part of circle)
Central angle vs circle angle
∠ACB = 90°
Angle in a semicircle
∠A + ∠C = 180°, ∠B + ∠D = 180°
Opposite angles of cyclic quadrilateral
exterior angle = interior opposite angle
Exterior angle property

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Circle terms easy

What is the longest chord of a circle called?

Q2 Perpendicular from centre to a chord medium

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

Q3 Perpendicular from centre to a chord medium

A chord of length 10 cm is at a distance of 12 cm from the centre of a circle. Find the radius.

Q4 Angle subtended by an arc medium

An arc of a circle subtends an angle of 80° at the centre. What angle does it subtend at a point on the major arc?

Q5 Angle in a semicircle hard

AB is a diameter of a circle with centre O, and C is a point on the circle such that angle OCA = 35°. Find angle OCB.

Q6 Cyclic quadrilateral medium

In a cyclic quadrilateral, angle A = 3x° and angle C = (x + 40)°. Find x.

Q7 Circle basics easy

What is the area of a circle of radius 14 cm? (Use π = 22/7)

Q8 Angle subtended by an arc easy

What is true about angles subtended by the same arc in the same segment of a circle?

Q9 Cyclic quadrilateral easy

ABCD is a cyclic quadrilateral. If angle B = 70°, what is angle D?

Q10 Circle terms easy

The region between a chord and its corresponding arc is called a:

Q11 Perpendicular from centre to a chord medium

A chord PQ of a circle with centre O has a perpendicular distance of 0 cm from O. What can be said about PQ?

Q12 Cyclic quadrilateral easy

In a cyclic quadrilateral, if one angle is 90°, what is its opposite angle?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

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.Perpendicular from centre / common chord

Let O and O′ be the centres of the circles with radii OA = 5 cm and O′A = 3 cm, and OO′ = 4 cm. Let the common chord be AB, meeting OO′ at M, with OM = x.

In right triangle OMA: AM² = OA² − OM² = 25 − x²

In right triangle O′MA: AM² = O′A² − O′M² = 9 − (4 − x)²

Equating: 25 − x² = 9 − (16 − 8x + x²) = −7 + 8x − x²

25 = −7 + 8x, so 8x = 32 and x = 4

AM² = 25 − 16 = 9, so AM = 3 cm

The common chord AB = 2 × AM = 2 × 3 = 6 cm.

2 AB and CD are two parallel chords of a circle of radius 5 cm, lying on opposite sides of the centre O. If AB = 8 cm and CD = 6 cm, find the distance between the chords.Equal/parallel chords and distance from centre

For chord AB: half of AB = 4 cm. Using OA² = OM² + AM²: 5² = OM² + 4², so 25 = OM² + 16, giving OM² = 9 and OM = 3 cm.

For chord CD: half of CD = 3 cm. Using OC² = ON² + CN²: 5² = ON² + 3², so 25 = ON² + 9, giving ON² = 16 and ON = 4 cm.

Since AB and CD are on opposite sides of the centre, the distance between them = OM + ON = 3 + 4 = 7 cm.

3 Prove that the perpendicular drawn from the centre of a circle to a chord bisects the chord.Perpendicular from centre bisects the chord

Given: A circle with centre O, and a chord AB. OM ⊥ AB, where M lies on AB.

To prove: AM = MB.

Proof: Join OA and OB. In triangles OMA and OMB:

  • OA = OB (radii of the same circle)
  • OM = OM (common side)
  • angle OMA = angle OMB = 90° (given OM ⊥ AB)

By RHS congruence, triangle OMA ≅ triangle OMB.

Therefore AM = MB (corresponding parts of congruent triangles), so OM bisects the chord AB.

4 Prove that a cyclic parallelogram is a rectangle.Cyclic quadrilateral

Given: ABCD is a cyclic parallelogram.

To prove: ABCD is a rectangle.

Proof: Since ABCD is cyclic, angle A + angle C = 180° (opposite angles of a cyclic quadrilateral).

Since ABCD is a parallelogram, opposite angles are equal, so angle A = angle C.

Substituting: angle A + angle A = 180°, so 2 × angle A = 180°, giving angle A = 90°.

Since one angle of the parallelogram is 90°, and opposite/adjacent angles of a parallelogram are related by co-interior angle sum (180°), all four angles are 90°.

A parallelogram with all angles equal to 90° is a rectangle. Hence ABCD is a rectangle.

5 PQ and RS are two parallel chords of a circle with centre O and radius 10 cm, such that PQ = 16 cm and RS = 12 cm. If PQ and RS are on opposite sides of O, find the distance between them.Perpendicular from centre to a chord

For chord PQ: half of PQ = 8 cm. OP² = OM² + PM²: 10² = OM² + 8², so 100 = OM² + 64, giving OM² = 36 and OM = 6 cm.

For chord RS: half of RS = 6 cm. OR² = ON² + RN²: 10² = ON² + 6², so 100 = ON² + 36, giving ON² = 64 and ON = 8 cm.

Since PQ and RS lie on opposite sides of the centre, the distance between them = OM + ON = 6 + 8 = 14 cm.

6 AB is a diameter of a circle, and C is a point on the circle such that angle CAB = 30°. Find angle ACB and angle ABC.Angle in a semicircle

Since AB is a diameter, the angle subtended by it at any point on the circle is a right angle (angle in a semicircle).

So angle ACB = 90°.

In triangle ABC, the sum of angles = 180°:

angle CAB + angle ABC + angle ACB = 180°

30° + angle ABC + 90° = 180°

angle ABC = 180° − 120° = 60°.

Hence angle ACB = 90° and angle ABC = 60°.

Previous-year board questions 4

Q1 In a figure, O is the centre of a circle and chord AB subtends an angle of 100° at the centre. Find angle ACB, where C is a point on the major arc. CBSE 2020 1 mark

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.

So angle AOB = 2 × angle ACB

100° = 2 × angle ACB

angle ACB = 50°.

Q2 AB is a chord of length 6 cm of a circle of radius 5 cm. Find the distance of the chord from the centre. CBSE 2019 2 marks

Let O be the centre and M be the midpoint of chord AB, so OM ⊥ AB and AM = 6/2 = 3 cm.

In right triangle OMA: OA² = OM² + AM²

5² = OM² + 3²

25 = OM² + 9

OM² = 16, so OM = 4 cm.

The chord is at a distance of 4 cm from the centre.

Q3 ABCD is a cyclic quadrilateral in which AB is parallel to CD. If angle B = 70°, find the remaining three angles of the quadrilateral. CBSE 2018 3 marks

Since ABCD is cyclic, angle B + angle D = 180° (opposite angles).

70° + angle D = 180°, so angle D = 110°.

Since AB ∥ CD, AD is a transversal, so angle A + angle D = 180° (co-interior angles).

angle A + 110° = 180°, so angle A = 70°.

Since ABCD is cyclic, angle A + angle C = 180°.

70° + angle C = 180°, so angle C = 110°.

Check: angle B + angle C = 70° + 110° = 180° (co-interior angles along BC, since AB ∥ CD) — consistent.

So angle A = 70°, angle C = 110°, angle D = 110° (and angle B = 70° given).

Q4 Prove that equal chords of a circle are equidistant from the centre. CBSE 2017 3 marks

Given: A circle with centre O, and two chords AB and CD such that AB = CD. OM ⊥ AB and ON ⊥ CD, where M lies on AB and N lies on CD.

To prove: OM = ON.

Proof: Since the perpendicular from the centre bisects the chord, M is the midpoint of AB and N is the midpoint of CD.

So AM = AB/2 and CN = CD/2. Since AB = CD (given), AM = CN.

Now consider right triangles OMA and ONC:

  • OA = OC (radii of the same circle)
  • AM = CN (proved above)
  • angle OMA = angle ONC = 90°

By RHS congruence, triangle OMA ≅ triangle ONC.

Therefore OM = ON (corresponding parts of congruent triangles, CPCT).

Hence equal chords AB and CD are equidistant from the centre O.

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