Class 9 — Mathematics (NCERT)

Circles

Class 9

✓By the end of this lesson students will be able to understand and apply properties of chords of a circle.
✓By the end of this lesson students will be able to relate angles subtended by an arc at the centre and at any point on the remaining part of the circle.
✓By the end of this lesson students will be able to identify and apply properties of cyclic quadrilaterals.
✓By the end of this lesson students will be able to solve problems involving chords, arcs, and angles in a circle.
✓By the end of this lesson students will be able to prove simple geometric results related to circles.

Key concepts

Chord of a Circle

A line segment joining any two points on a circle is called a chord of the circle. The longest chord of a circle is its diameter.

Theorem 10.1: Equal Chords Subtend Equal Angles at the Centre

If two chords of a circle are equal in length, then they subtend equal angles at the centre of the circle.

Theorem 10.2: Converse of Theorem 10.1

If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal in length.

Theorem 10.3: Perpendicular from Centre to Chord

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

Theorem 10.4: Converse of Theorem 10.3

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

Theorem 10.6: Equal Chords are Equidistant from the Centre

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

Theorem 10.7: Converse of Theorem 10.6

Chords equidistant from the centre of a circle are equal in length.

Angle Subtended by an Arc

An arc of a circle subtends an angle at the centre and also at any point on the remaining part of the circle. The angle is formed by joining the endpoints of the arc to the point.

Theorem 10.8: Angle at Centre is Double the Angle at Circumference

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.

∠AOB = 2 × ∠APB (where AOB is angle at centre, APB is angle at circumference by arc AB)

Theorem 10.9: Angles in the Same Segment

Angles in the same segment of a circle are equal. A special case is that the angle in a semicircle is a right angle (90°).

Cyclic Quadrilateral

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

Theorem 10.11: Opposite Angles of a Cyclic Quadrilateral

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

∠A + ∠C = 180°, ∠B + ∠D = 180° (for cyclic quadrilateral ABCD)

Theorem 10.12: Converse of Theorem 10.11

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

Key facts to remember

1The perpendicular from the centre of a circle to a chord bisects the chord.
2Equal chords of a circle subtend equal angles at the centre, and conversely.
3The 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.
4Angles in the same segment of a circle are equal.
5The angle in a semicircle is a right angle (90°).
6The sum of either pair of opposite angles of a cyclic quadrilateral is 180°, and conversely.

Worked examples

Example 1

In a circle with centre O, AB and CD are two parallel chords. If AB = 10 cm and CD = 24 cm, and the radius of the circle is 13 cm, find the distance between the chords.

IDraw a circle with centre O and radius 13 cm. Draw two parallel chords AB and CD. Draw OM ⊥ AB and ON ⊥ CD. Since AB || CD, M, O, N are collinear.

IIBy Theorem 10.3, the perpendicular from the centre to a chord bisects the chord. So, AM = MB = AB/2 = 10/2 = 5 cm and CN = ND = CD/2 = 24/2 = 12 cm.

IIIIn right-angled ΔOMA, OA is the hypotenuse (radius). By Pythagoras Theorem, OA² = OM² + AM².

IV13² = OM² + 5²

V169 = OM² + 25

VIOM² = 169 - 25 = 144

VIIOM = √144 = 12 cm.

VIIIIn right-angled ΔONC, OC is the hypotenuse (radius). By Pythagoras Theorem, OC² = ON² + CN².

913² = ON² + 12²

10169 = ON² + 144

11ON² = 169 - 144 = 25

12ON = √25 = 5 cm.

13Since the chords are of different lengths, they must lie on opposite sides of the centre O for them to be parallel and distinct. Therefore, the distance between the chords is MN = OM + ON.

14MN = 12 cm + 5 cm = 17 cm.

Answer

The distance between the chords is 17 cm.

If the problem does not specify the relative positions of parallel chords, it is generally assumed they are on opposite sides of the centre if their lengths are different.

Example 2

In Fig., A, B, C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

IFirst, find the angle subtended by arc AC at the centre O. ∠AOC = ∠AOB + ∠BOC.

II∠AOC = 60° + 30° = 90°.

IIIThe arc AC subtends ∠AOC at the centre and ∠ADC at point D on the remaining part of the circle.

IVBy Theorem 10.8, 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.

VTherefore, ∠AOC = 2 × ∠ADC.

VISubstitute the value of ∠AOC: 90° = 2 × ∠ADC.

VII∠ADC = 90° / 2 = 45°.

Answer

∠ADC = 45°.

Example 3

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70° and ∠BAC = 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

IGiven ABCD is a cyclic quadrilateral.

IIAngles in the same segment of a circle are equal. Arc CD subtends ∠CAD and ∠DBC. Thus, ∠CAD = ∠DBC = 70°.

IIINow, find ∠DAB. ∠DAB = ∠DAC + ∠CAB = 70° + 30° = 100°.

IVIn a cyclic quadrilateral, the sum of opposite angles is 180° (Theorem 10.11). Therefore, ∠DAB + ∠BCD = 180°.

VSubstitute the value of ∠DAB: 100° + ∠BCD = 180°.

VI∠BCD = 180° - 100° = 80°.

VIINow, for the second part, given AB = BC. In ΔABC, since AB = BC, the angles opposite to these sides are equal.

VIIISo, ∠BCA = ∠BAC = 30°.

9We need to find ∠ECD. We know ∠BCD = 80° and ∠BCA = 30°.

10∠ECD = ∠BCD - ∠BCA = 80° - 30° = 50°.

Answer

∠BCD = 80°, ∠ECD = 50°.

Common mistakes

✗Confusing the angle subtended at the centre with the angle subtended at the circumference, often forgetting to double or halve the angle correctly.
✗Incorrectly assuming a line from the centre to a chord is perpendicular just because it bisects the chord, or vice-versa, without explicitly stating the relevant theorem.
✗Applying properties of cyclic quadrilaterals to any quadrilateral, instead of only those whose all four vertices lie on a circle.
✗Not correctly identifying the arc or segment when applying theorems related to angles in the same segment or angle at the centre.
✗Assuming that if a quadrilateral has one pair of opposite angles summing to 180°, the other pair also sums to 180° without first establishing that the quadrilateral is cyclic.

Exam tips

★Always draw a neat and labelled diagram for each problem. This helps in visualising the given information and the required solution.
★Clearly state the theorems or properties of circles being used at each step of your solution. For example, 'Angles in the same segment are equal' or 'The sum of opposite angles of a cyclic quadrilateral is 180°'.
★Break down complex problems into smaller, manageable steps. Solve one part at a time to avoid errors.
★Pay close attention to the wording of the question, especially regarding major/minor arcs or specific points on the circle, as this can affect the application of theorems.
★Practice solving a variety of problems from your NCERT textbook and reference books to become familiar with different applications of the theorems.

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