Class 9 Maths Chapter 9: Circles

Circles is Chapter 9 of NCERT Class 9 Maths. It covers the theorems that link chords, arcs and angles in a circle, from equal chords and perpendiculars drawn from the centre to angles in the same segment and cyclic quadrilaterals. Watch the five video lessons below, then revise with these notes, theorems and solved examples.

What you will learn

• How equal chords relate to the angles they subtend at the centre
• The perpendicular from the centre to a chord and how to find chord lengths
• Why equal chords are equidistant from the centre
• The relation between the angle at the centre and the angle on the circle
• Properties of cyclic quadrilaterals and how to use them in problems

Video Lessons

Circles - Part 1

Circles - Part 2

Circles - Part 3

Circles - Part 4

Circles - Part 5

Key Concepts and Formulas

Basic Terms Recap

A circle is the set of all points in a plane at a fixed distance (the radius) from a fixed point (the centre). Knowing these words clearly makes the theorems much easier to follow.

• Chord: a line segment joining any two points on the circle. The diameter is the longest chord.
• Arc: a piece of the circle. The smaller piece is the minor arc and the larger one is the major arc.
• Segment: the region between a chord and an arc. Sector: the region between an arc and two radii.
• Congruent circles have equal radii.

Equal Chords and Angles at the Centre

These results also hold for chords of congruent circles.

AB = CD ⇒ ∠AOB = ∠COD
∠AOB = ∠COD ⇒ AB = CD

• Equal chords of a circle subtend equal angles at the centre.
• Converse: if the angles subtended by two chords at the centre are equal, the chords are equal.
• Equal chords cut off congruent arcs, and congruent arcs have equal chords.

Perpendicular from the Centre to a Chord

Drawing the perpendicular from the centre creates a right triangle with the radius as hypotenuse, which is how most chord-length questions are solved.

OM ⊥ AB ⇒ AM = MB
(half chord)² + d² = r²
Chord length = 2√(r² − d²)
where r = radius, d = distance from centre

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

Equal Chords and Their Distances from the Centre

The distance of a chord from the centre always means the length of the perpendicular from the centre to the chord.

AB = CD ⇔ OM = ON
(OM, ON are perpendiculars to AB, CD)

• Equal chords are equidistant from the centre.
• Converse: chords equidistant from the centre are equal.
• From the chord-length formula, a longer chord is nearer to the centre.

Angle Subtended by an Arc

Most angle questions in this chapter use one of these three results.

∠AOB = 2 × ∠ACB
(C on the remaining part of the circle)
Same segment: ∠ACB = ∠ADB
Angle in a semicircle = 90°

• 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 are equal.
• Converse: if a segment joining two points subtends equal angles at two other points on the same side of it, all four points lie on a circle (they are concyclic).

Cyclic Quadrilaterals

A quadrilateral whose four vertices all lie on one circle is called a cyclic quadrilateral.

∠A + ∠C = 180°
∠B + ∠D = 180°
Exterior angle = interior opposite angle

• The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
• Converse: if a pair of opposite angles of a quadrilateral adds up to 180°, the quadrilateral is cyclic.

Solved Examples

Example 1: A chord is 5 cm from the centre of a circle of radius 13 cm. Find its length.

Half chord = √(r² − d²)
= √(13² − 5²) = √(169 − 25)
= √144 = 12 cm
Chord = 2 × 12
∴ Chord = 24 cm

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

Half chord = 16 ÷ 2 = 8 cm
d = √(10² − 8²) = √(100 − 64)
= √36
∴ d = 6 cm

Example 3: An arc subtends an angle of 128° at the centre. Find the angle it subtends at a point on the remaining part of the circle.

Angle on circle = ½ × angle at centre
= ½ × 128°
∴ Angle = 64°

Example 4: ABCD is a cyclic quadrilateral with ∠A = 74° and ∠B = 101°. Find ∠C and ∠D.

∠A + ∠C = 180° ⇒ ∠C = 180° − 74° = 106°
∠B + ∠D = 180° ⇒ ∠D = 180° − 101° = 79°
∴ ∠C = 106°, ∠D = 79°

Example 5: AB is a diameter of a circle and C is a point on the circle with ∠CAB = 37°. Find ∠ABC.

∠ACB = 90° (angle in a semicircle)
∠ABC = 180° − 90° − 37°
∴ ∠ABC = 53°

Example 6: A, B, C and D lie on a circle with centre O, and C and D are on the same side of AB. If ∠ACB = 42°, find ∠ADB and ∠AOB.

∠ADB = ∠ACB (same segment) = 42°
∠AOB = 2 × ∠ACB = 2 × 42°
∴ ∠ADB = 42°, ∠AOB = 84°

Example 7: Two parallel chords of lengths 6 cm and 8 cm lie on opposite sides of the centre of a circle of radius 5 cm. Find the distance between them.

d₁ = √(5² − 3²) = √(25 − 9) = 4 cm
d₂ = √(5² − 4²) = √(25 − 16) = 3 cm
Opposite sides ⇒ distance = d₁ + d₂
∴ Distance = 4 + 3 = 7 cm

If the two chords were on the same side of the centre, the gap would be 4 − 3 = 1 cm instead.

Common Mistakes to Avoid

• Using the "angle at centre is double" theorem with a point on the wrong arc. The point must be on the remaining part of the circle, not on the arc itself.
• In a cyclic quadrilateral, adding adjacent angles to get 180°. Only opposite angles are supplementary.
• Finding the half chord with √(r² − d²) and then forgetting to double it for the full chord.
• Using "angle in a semicircle is 90°" when the chord is not a diameter. Check that the chord passes through the centre first.

Quick Revision

Chord = 2√(r² − d²)
Angle at centre = 2 × angle on circle
Angle in a semicircle = 90°
Cyclic quad: ∠A + ∠C = ∠B + ∠D = 180°

• Equal chords ↔ equal angles at the centre ↔ equal distances from the centre
• The perpendicular from the centre bisects a chord
• Angles in the same segment are equal

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