K2KARAKORAM
मराठीEnglishहिंदी
Open the app

CBSE · Class 9 · Mathematics

Chapter 5: Circles

6 concepts · 48 practice questions with answers · Free to read

This lesson is also in: मराठी · हिंदी

Parts of a circle and circles through points

From daily life

A wheel has a centre, spokes (radii) and a rim. A bangle is a circle drawn through many points at the same distance from the centre.

Simple definition

A circle is all points at a fixed distance (radius) from a fixed point (centre). A chord joins two points on it; the longest chord is the diameter = 2 × radius. A piece of the circle is an arc. Through 3 non-collinear points there is exactly one circle.

Formula: diameter = 2r; 1 circle through 3 non-collinear points

Odiameterrchord

In the figure: O is the centre, r the radius, the longest chord through O is the diameter.

Rule 1: Count the circles — Through one point: infinitely many. Through two points: infinitely many. Through three non-collinear points: exactly one. Through three collinear points: none.

  • Easy: Points A, B, C not on a line: one circle.
  • Medium: Points P, Q, R on one line: no circle.
  • Exam-type: Two points: centre lies on the perpendicular bisector of the segment.

Rule 2: Radius and diameter — All radii of a circle are equal; the diameter is twice the radius.

  • Easy: Radius 7 cm: diameter 14 cm.
  • Medium: Diameter 18 cm: radius 9 cm.
  • Exam-type: Longest chord of a circle with radius 5 cm is 10 cm.
Remember

A circle has only one centre and infinitely many diameters.

Common mistake

Thinking the diameter equals the radius, or that a chord cannot pass through the centre.

Memory clue

"Diameter is the double-radius chord."

Practice questions

Easy
  1. 1The radius of a circle is 19 cm. What is its diameter?
  2. 2The longest chord of a circle is 28 cm. What is the radius of the circle?
  3. 3How many circles can pass through three collinear points and no other condition?
Medium
  1. 4How many circles can pass through two given distinct points?
  2. 5A circle has centre O and radius 9 cm. A point P is 13 cm from O. Where is P?
  3. 6The perpendicular bisectors of chords AB and BC of a circle meet at O, and OA = 12 cm. What is the length of OC?
Challenging
  1. 7The perpendicular bisectors of chords AB and BC of a circle meet at O, and OA = 13 cm. What is the length of OC?
  2. 8Two circles have the same centre. The radii are 11 cm and 9 cm. What is the width of the ring between them?
Show answers (try them yourself first)
  1. 138 cm
  2. 214 cm
  3. 3none
  4. 4infinitely many
  5. 5outside the circle
  6. 612 cm
  7. 713 cm
  8. 82 cm

Chords and the angles they subtend at the centre

From daily life

The hands of a clock make a bigger angle at the centre when the tips are further apart. A longer chord opens a bigger angle at the centre.

Simple definition

In one circle (or equal circles), equal chords subtend equal angles at the centre. Conversely, chords that subtend equal angles at the centre are equal.

Formula: AB = CD ⇔ ∠AOB = ∠COD

ABCD

In the figure: Chords AB and CD are equal, so triangles AOB and COD are congruent and ∠AOB = ∠COD.

Rule 1: Equal chords, equal angles — Use SSS: two radii and the chord form congruent triangles.

  • Easy: AB = CD, ∠AOB = 70° ⇒ ∠COD = 70°.
  • Medium: ∠POQ = ∠ROS ⇒ PQ = RS.
  • Exam-type: Two equal chords of a circle give equal angles; a bigger chord gives a bigger angle.

Rule 2: Find an unknown angle — In the isosceles triangle AOB, base angles are equal and all three angles add to 180°.

  • Easy: ∠AOB = 80° ⇒ ∠OAB = (180° − 80°)/2 = 50°.
  • Medium: ∠OAB = 35° ⇒ ∠AOB = 110°.
  • Exam-type: Equilateral chord (AB = r): ∠AOB = 60°.
Remember

Chord length and the angle at the centre grow together.

Common mistake

Using the result for two different circles of different radii.

Memory clue

"Equal rope, equal opening."

Practice questions

Easy
  1. 1In a circle with centre O, chords AB and CD are equal and ∠AOB = 140°. What is ∠COD?
  2. 2In a circle with centre O, chords AB and CD are equal and ∠AOB = 34°. What is ∠COD?
  3. 3In a circle with centre O, chord AB subtends ∠AOB = 30° at the centre. Find ∠OAB.
Medium
  1. 4A chord of a circle subtends 120° at the centre. How many chords of this same length can be placed end to end around the whole circle?
  2. 5A chord of a circle subtends 60° at the centre. How many chords of this same length can be placed end to end around the whole circle?
  3. 6Chords AB, BC and CD are equal, and each subtends 25° at the centre O. Find ∠AOD.
Challenging
  1. 7Chords AB, BC and CD are equal, and each subtends 20° at the centre O. Find ∠AOD.
  2. 8In triangle OAB, O is the centre of a circle and A, B lie on it. If ∠OAB = 35°, find ∠AOB.
Show answers (try them yourself first)
  1. 1140°
  2. 234°
  3. 375°
  4. 43
  5. 56
  6. 675°
  7. 760°
  8. 8110°

Perpendicular from the centre to a chord

From daily life

Folding a round paper along a chord, the fold from the centre meets it at its middle point at a right angle.

Simple definition

The perpendicular from the centre to a chord bisects the chord. Conversely, the line from the centre to the midpoint of a chord is perpendicular to it. Using Pythagoras: r² = d² + (c/2)².

Formula: r² = d² + (c/2)²

ABMO

In the figure: OM ⟂ AB, so AM = MB. Triangle OMA is right-angled with hypotenuse OA = r.

Rule 1: Find the chord — Half-chord = √(r² − d²); the full chord is twice that.

  • Easy: r = 5, d = 3: half-chord 4, chord 8.
  • Medium: r = 13, d = 5: chord 24.
  • Exam-type: r = 10, d = 6: chord 16.

Rule 2: Find the distance or radius — Draw OM and use the right triangle.

  • Easy: Chord 16, r = 10: d = √(100 − 64) = 6.
  • Medium: Chord 24, d = 5: r = √(144 + 25) = 13.
  • Exam-type: Chord 6, d = 4: r = 5.
Remember

A chord through the centre has d = 0 and length 2r.

Common mistake

Using the whole chord instead of half of it in Pythagoras.

Memory clue

"Half the chord, then Pythagoras."

Practice questions

Easy
  1. 1In a circle with centre O, OM is perpendicular to chord AB, and AB = 14 cm. Find AM.
  2. 2In a circle with centre O, OM is perpendicular to chord AB, and AB = 20 cm. Find AM.
  3. 3The perpendicular from the centre O of a circle meets chord AB at M, and AM = 7 cm. Find AB.
Medium
  1. 4A circle has radius 20 cm. A chord is 12 cm from the centre. Find the length of the chord.
  2. 5A chord of length 30 cm is drawn in a circle of radius 17 cm. How far is the chord from the centre?
  3. 6A chord of length 40 cm is drawn in a circle of radius 25 cm. How far is the chord from the centre?
Challenging
  1. 7A chord of length 16 cm is 15 cm away from the centre of a circle. Find the radius.
  2. 8Two circles have the same centre and radii 17 cm and 15 cm. A chord of the larger circle touches the smaller circle. Find the length of the chord.
Show answers (try them yourself first)
  1. 17 cm
  2. 210 cm
  3. 314 cm
  4. 432 cm
  5. 58 cm
  6. 615 cm
  7. 717 cm
  8. 816 cm

Distance of chords from the centre

From daily life

In a round cake, a cut close to the middle is long; a cut near the edge is short. Equal cuts are equally far from the middle.

Simple definition

Equal chords are equidistant from the centre, and chords equidistant from the centre are equal. The longer the chord, the closer it is to the centre.

Formula: c₁ = c₂ ⇔ d₁ = d₂; bigger c ⇒ smaller d

dd

In the figure: Both red chords are equal, so both green distances from O are equal.

Rule 1: Compare chords — Compare distances: the smaller distance means the longer chord.

  • Easy: d₁ = 3, d₂ = 5 (same circle): chord 1 is longer.
  • Medium: Equal chords: both are 4 cm from centre.
  • Exam-type: The diameter is nearest to the centre (d = 0) and longest.

Rule 2: Use the formula — Same radius means r² is common, so c₁²/4 + d₁² = c₂²/4 + d₂².

  • Easy: Chord 8 at d = 3: r = 5. A chord at d = 4 has length 6.
  • Medium: Two equal chords of 10 cm in a circle of r = 13 are each 12 cm from the centre.
  • Exam-type: Chord 14, r = 25: d = 24.
Remember

Distance means the perpendicular length from the centre.

Common mistake

Saying the longer chord is farther from the centre.

Memory clue

"Big chord hugs the centre."

Practice questions

Easy
  1. 1Chords AB and CD of a circle are equal. If the distance of AB from the centre is 6 cm, what is the distance of CD from the centre?
  2. 2Chords AB and CD of a circle are equal. If the distance of AB from the centre is 11 cm, what is the distance of CD from the centre?
  3. 3Two chords of a circle have lengths 37 cm and 23 cm. Which chord is nearer to the centre?
Medium
  1. 4In a circle of radius 26 cm, two equal chords each of length 48 cm are drawn. Find the distance of each chord from the centre.
  2. 5Two chords of a circle are each 4 cm from the centre. One of them is 20 cm long. What is the length of the other?
  3. 6Two chords of a circle are each 4 cm from the centre. One of them is 11 cm long. What is the length of the other?
Challenging
  1. 7In a circle of radius 25 cm, two parallel chords of lengths 48 cm and 14 cm lie on opposite sides of the centre. What is the distance between them?
  2. 8In a circle of radius 17 cm, two parallel chords of lengths 16 cm and 30 cm lie on the same side of the centre. What is the distance between them?
Show answers (try them yourself first)
  1. 16 cm
  2. 211 cm
  3. 3the chord of length 37 cm
  4. 410 cm
  5. 520 cm
  6. 611 cm
  7. 731 cm
  8. 87 cm
Live figure

Distance of chords from the centre

In the app you can move this figure with your finger.

Open in the app

Angles subtended by an arc

From daily life

A photographer standing anywhere on a curved stage edge sees the same two lights at the same angle. The camera at the centre sees them twice as wide.

Simple definition

The angle an arc makes at the centre is twice the angle it makes at any point on the remaining circle. Angles in the same segment are equal, and the angle in a semicircle is 90°.

Formula: ∠AOB = 2∠ACB; angle in semicircle = 90°

ABC2xx

In the figure: ∠AOB = 2x at the centre and ∠ACB = x on the circle, both standing on arc AB.

Rule 1: Centre and circumference — Circumference angle = half of the centre angle on the same arc.

  • Easy: Centre angle 100° ⇒ ACB = 50°.
  • Medium: ∠ACB = 35° ⇒ ∠AOB = 70°.
  • Exam-type: Angle in a semicircle: 90°.

Rule 2: Same segment — Points on the same side of the chord see it at equal angles.

  • Easy: ∠ACB = 40° ⇒ ∠ADB = 40° if C, D lie on the same side.
  • Medium: If AB is a diameter, ∠ACB = 90°.
  • Exam-type: Chord PQ subtends 60° at the circumference: angle at the centre is 120°.
Remember

The theorem uses the arc that does not contain the point on the circle.

Common mistake

Halving when you should double, e.g. centre angle taken as half the circumference angle.

Memory clue

"Centre is the boss: double."

Practice questions

Easy
  1. 1An arc of a circle subtends an angle of 50° at the centre. What angle does it subtend at a point on the remaining part of the circle?
  2. 2An arc of a circle subtends an angle of 67° at a point on the circle (on the major arc). What angle does it subtend at the centre?
  3. 3AB is a diameter of a circle and C is a point on the circle. If ∠CAB = 27°, find ∠CBA.
Medium
  1. 4A, B, C and D lie on a circle with C and D on the same side of chord AB. If ∠ACB = 27°, find ∠ADB.
  2. 5Chord AB subtends ∠AOB = 130° at the centre O. A point Q lies on the minor arc AB. Find ∠AQB.
  3. 6A point P on a circle sees chord AB at an angle of 69°. What is the reflex angle AOB at the centre O (on the same side as the minor arc)?
Challenging
  1. 7In a circle with centre O, ∠OAB = 60° where A, B lie on the circle. A point P lies on the major arc AB. Find ∠APB.
  2. 8A triangle ABC is inscribed in a circle with AB as a diameter, and ∠CAB = 64°. Find ∠CBA.
Show answers (try them yourself first)
  1. 125°
  2. 2134°
  3. 363°
  4. 427°
  5. 5115°
  6. 6222°
  7. 730°
  8. 826°
Live figure

Angles subtended by an arc

In the app you can move this figure with your finger.

Open in the app

Cyclic quadrilaterals

From daily life

A quadrilateral whose four corners lie on a round bangle is "cyclic". Opposite corners then always share 180°.

Simple definition

In a cyclic quadrilateral all four vertices lie on a circle. Opposite angles are supplementary (sum 180°) and an exterior angle equals the interior opposite angle. Conversely, if opposite angles sum to 180°, the quadrilateral is cyclic.

Formula: ∠A + ∠C = 180°, ∠B + ∠D = 180°

ABCDA + C = 180

In the figure: Opposite corners A and C add to 180°, and so do B and D.

Rule 1: Find the angle — Opposite angle = 180° minus the given one.

  • Easy: ∠A = 85° ⇒ ∠C = 95°.
  • Medium: ∠B = 110° ⇒ ∠D = 70°.
  • Exam-type: Exterior angle at D = 75° ⇒ ∠B = 75°.

Rule 2: Is it cyclic? — Check whether a pair of opposite angles sums to 180°.

  • Easy: Angles 80°, 100°, 100°, 80° in order: 80 + 100 = 180 (opposite: 80 & 100), yes.
  • Medium: Rectangle: opposite 90° + 90° = 180°, always cyclic.
  • Exam-type: A parallelogram that is not a rectangle is not cyclic.
Remember

Every rectangle and square is cyclic. Every cyclic parallelogram is a rectangle.

Common mistake

Adding adjacent angles instead of opposite ones.

Memory clue

"Opposite pairs make a straight line: 180°."

Practice questions

Easy
  1. 1ABCD is a cyclic quadrilateral with ∠A = 97°. Find ∠C.
  2. 2ABCD is a cyclic quadrilateral. Side AB is extended to E and ∠CBE = 115°. Find ∠ADC.
  3. 3ABCD is an isosceles trapezium with AB ∥ DC, so it is cyclic. If ∠A = 62°, find ∠C.
Medium
  1. 4In a quadrilateral ABCD, ∠A = 74° and ∠C = 106°. Is it a cyclic quadrilateral (judging by this pair of angles)?
  2. 5In a cyclic quadrilateral ABCD, ∠A = (2x + 10)° and ∠C = (2x + 70)°. Find x.
  3. 6The angles A, B, C of a cyclic quadrilateral ABCD are in the ratio 3 : 4 : 6. Find ∠D.
Challenging
  1. 7Can an isosceles trapezium be inscribed in a circle (be cyclic)?
  2. 8ABCD is a cyclic quadrilateral. Diagonal BD makes ∠ADB = 54° and ∠BDC = 25°. Find ∠ABC.
Show answers (try them yourself first)
  1. 183°
  2. 2115°
  3. 3118°
  4. 4Yes, it is cyclic
  5. 525
  6. 6100°
  7. 7Yes
  8. 8101°