Circles — Class 9 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Circles" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Circles" — 8 important questions with detailed answers for CBSE board exam preparat…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 9 Mathematics — Circles. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Circles — All 8 Questions Solved
Q1. Define a circle and state the following properties: (1) All radii of a circle are equal. (2) A chord is a line segment joining any two points on the circle. (3) The longest chord is the diameter. If a circle has radius 7 cm, find the diameter and the circumference.
Step 1: Definition of Circle: A circle is the locus of all points in a plane that are at a constant distance (radius) from a fixed point (center).
Step 2: Properties:
(1) All radii of a circle are equal: Since all points on the circle are equidistant from the center, all radii r are equal.
(2) Chord: A line segment with both endpoints on the circle.
Examples: diameter (through center), any other line segment between two points on circle.
(3) Diameter is the longest chord: Diameter = 2r (passes through center).
Any other chord < diameter (doesn't pass through center).
Step 3: Given: radius r = 7 cm
Step 4: Calculate diameter:
Diameter d = 2r = 2 × 7 = 14 cm
Step 5: Calculate circumference (perimeter of circle):
Circumference C = 2πr
C = 2 × π × 7
C = 14π cm
C ≈ 14 × 3.14159
C ≈ 43.98 cm or approximately 44 cm
Step 6: Alternative circumference formula:
C = πd = π × 14 = 14π cm ✓
Answer: Diameter = 14 cm; Circumference = 14π cm ≈ 43.98 cm.
Q2. State the theorem about angles subtended by an arc at the center and on the circumference: If an arc AB subtends an angle of 60° at the center O, what angle does the same arc subtend at a point C on the circumference (on the major arc)?
Step 1: Theorem: The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the circle (on the major arc).
Angle at center = 2 × Angle at circumference
Step 2: Given:
Arc AB subtends angle AOB = 60° at center O
Point C is on the major arc AB
Find: angle ACB (angle at circumference)
Step 3: Apply the theorem:
angle AOB = 2 × angle ACB
60° = 2 × angle ACB
angle ACB = 60°/2
angle ACB = 30°
Step 4: Verification: This theorem holds for all positions of point C on the major arc. If C moves along the major arc, angle ACB remains 30°.
Step 5: Special case - Angle in a semicircle:
If arc AB is a diameter (semicircle), angle at center = 180°
Angle at circumference = 180°/2 = 90°
This means any angle inscribed in a semicircle is a right angle.
Answer: The angle subtended at the circumference (point C) = 30°.
Q3. In circle with center O and radius 5 cm, a chord AB is at a perpendicular distance of 3 cm from the center O. Find the length of chord AB.
Step 1: Given:
Circle with center O and radius r = 5 cm
Chord AB at perpendicular distance d = 3 cm from center O
Step 2: Let M be the foot of the perpendicular from O to chord AB.
Then OM ⊥ AB and OM = 3 cm
Step 3: Important property: The perpendicular from the center of a circle to a chord bisects the chord.
Therefore: AM = MB = AB/2
Step 4: In right triangle OMA:
angle OMA = 90°
OA = 5 cm (radius, hypotenuse)
OM = 3 cm (perpendicular distance)
AM = ? (half of chord)
Step 5: By Pythagoras theorem:
OA² = OM² + AM²
5² = 3² + AM²
25 = 9 + AM²
AM² = 25 - 9
AM² = 16
AM = 4 cm
Step 6: Since M bisects the chord:
AB = 2 × AM
AB = 2 × 4
AB = 8 cm
Step 7: Verification using right triangle OMB:
OB² = OM² + MB²
5² = 3² + 4²
25 = 9 + 16
25 = 25 ✓
Answer: Length of chord AB = 8 cm.
Q4. State the theorem about angles in the same segment: Two angles subtended by the same chord at two different points on the same arc are equal. In circle with center O, chord AB subtends angle 40° at point C on the major arc. Find the angle it subtends at point D, also on the major arc.
Step 1: Theorem (Angles in the Same Segment): Angles subtended by the same chord at different points on the same arc of a circle are equal.
If chord AB subtends angles at points C and D on the same arc, then angle ACB = angle ADB.
Step 2: Given:
Circle with center O
Chord AB
Point C on major arc: angle ACB = 40°
Point D also on major arc
Find: angle ADB
Step 3: Since C and D are both on the major arc (same segment), and they are both subtended by the same chord AB:
angle ADB = angle ACB
Step 4: Therefore:
angle ADB = 40°
Step 5: Proof using central angle:
Let angle AOB (central angle) = θ
Angle at circumference = θ/2
Since both C and D see the same central angle AOB:
angle ACB = θ/2
angle ADB = θ/2
Hence angle ACB = angle ADB
Step 6: Different segments give different angles:
If point E were on the minor arc, angle AEB would be supplementary to angle ACB.
angle AEB = 180° - 40° = 140°
Answer: Angle ADB = 40° (same as angle ACB, being on the same arc).
Q5. In a circle, two chords AB and CD intersect at point P inside the circle. If AP = 6 cm, PB = 4 cm, and CP = 3 cm, find the length of PD.
Step 1: Theorem (Intersecting Chords): When two chords intersect inside a circle, the products of their segments are equal.
If chords AB and CD intersect at P, then: AP × PB = CP × PD
Step 2: Given:
Chord AB with intersection point P
AP = 6 cm
PB = 4 cm
Chord CD with intersection point P
CP = 3 cm
PD = ?
Step 3: Apply the intersecting chords theorem:
AP × PB = CP × PD
6 × 4 = 3 × PD
24 = 3 × PD
Step 4: Solve for PD:
PD = 24/3
PD = 8 cm
Step 5: Verification:
Product of segments of chord AB = AP × PB = 6 × 4 = 24
Product of segments of chord CD = CP × PD = 3 × 8 = 24 ✓
Step 6: Therefore:
Chord AB = AP + PB = 6 + 4 = 10 cm
Chord CD = CP + PD = 3 + 8 = 11 cm
Answer: Length of PD = 8 cm.
Q6. A chord of a circle subtends angles of 60° at two points on the major arc and 120° at the minor arc. Verify this using the angle-in-segment theorem.
Step 1: Angle-in-Segment Theorem Verification:
Angles subtended by a chord at points on the major arc are equal.
Angles subtended by a chord at points on the minor arc are equal.
Angles in the major arc + angles in the minor arc = 180°
Step 2: Given:
Chord AB
Angle at point C on major arc = 60°
Angle at point D on minor arc = 120°
Step 3: Verify the relationship:
angle ACB + angle ADB = 60° + 120° = 180° ✓
Step 4: Explanation using central angle:
Let the central angle AOB = θ
For point C on major arc:
angle ACB = θ/2 = 60°
Therefore: θ = 120°
Step 5: For point D on minor arc:
The angle subtended at a point on the minor arc is:
angle ADB = (360° - θ)/2 = (360° - 120°)/2 = 240°/2 = 120°
Step 6: Alternatively:
angle ACB + angle ADB = θ/2 + (360° - θ)/2
= [θ + 360° - θ]/2
= 360°/2
= 180° ✓
Step 7: This confirms:
Angles subtended by a chord at points on opposite arcs are supplementary (sum to 180°).
Answer: The theorem is verified: angles in major arc (60°) + angles in minor arc (120°) = 180°.
Q7. In a circle with center O and radius 6 cm, an arc AB subtends an angle of 90° at the center. Find the length of arc AB and the area of the sector AOB.
Step 1: Given:
Circle with center O and radius r = 6 cm
Central angle subtended by arc AB = 90°
Step 2: Arc Length Formula:
Arc length = (θ/360°) × 2πr
Where θ is the central angle in degrees.
Step 3: Calculate arc length:
Arc AB = (90°/360°) × 2π × 6
Arc AB = (1/4) × 2π × 6
Arc AB = (1/4) × 12π
Arc AB = 3π cm
Arc AB ≈ 3 × 3.14159 ≈ 9.42 cm
Step 4: Sector Area Formula:
Area of sector = (θ/360°) × πr²
Where θ is the central angle in degrees.
Step 5: Calculate sector area:
Area of sector AOB = (90°/360°) × π × 6²
Area of sector AOB = (1/4) × π × 36
Area of sector AOB = 9π cm²
Area of sector AOB ≈ 9 × 3.14159 ≈ 28.27 cm²
Step 6: Verification:
Full circle: arc = 2π × 6 = 12π cm, area = π × 6² = 36π cm²
Quarter circle: arc = 12π/4 = 3π cm ✓, area = 36π/4 = 9π cm² ✓
Answer: Arc length AB = 3π cm ≈ 9.42 cm; Area of sector AOB = 9π cm² ≈ 28.27 cm².
Q8. In a cyclic quadrilateral ABCD, angle A = 75° and angle C = 105°. Find angles B and D. Verify that opposite angles sum to 180°.
Step 1: Cyclic Quadrilateral: A quadrilateral whose all four vertices lie on a circle.
Step 2: Property of Cyclic Quadrilateral: The opposite angles of a cyclic quadrilateral are supplementary.
angle A + angle C = 180°
angle B + angle D = 180°
Step 3: Given:
angle A = 75°
angle C = 105°
Step 4: Verify the sum of given angles:
angle A + angle C = 75° + 105° = 180° ✓
This confirms ABCD is a cyclic quadrilateral (though the problem states it).
Step 5: Since opposite angles sum to 180°:
angle A + angle C = 180° ✓
angle B + angle D = 180° ... (1)
Step 6: We need another condition to find B and D individually. However, from the property of cyclic quadrilaterals and the sum of all angles = 360°:
angle A + angle B + angle C + angle D = 360°
75° + angle B + 105° + angle D = 360°
angle B + angle D = 360° - 180°
angle B + angle D = 180° ✓
Step 7: Without additional information, B and D cannot be uniquely determined individually. But we know:
angle B + angle D = 180°
angle B = 180° - angle D
angle D = 180° - angle B
Step 8: Example: If angle B = 85°, then angle D = 95°.
Verification: 75° + 85° + 105° + 95° = 360° ✓
Answer: angle B + angle D = 180°. Opposite angles are supplementary: angle A + angle C = 180° and angle B + angle D = 180°.
Circles — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. A line segment joining two points on a circle is called a:
- A. Radius
- B. Diameter
- C. Chord
- D. Arc
Answer: C. Chord — A chord is defined as a line segment whose endpoints both lie on the circumference of a circle.
MCQ 2. The perpendicular from the center of a circle to a chord ____ the chord.
- A. bisects
- B. trisects
- C. passes through the midpoint
- D. doubles
Answer: A. bisects — This is a fundamental theorem of circles, stating that a line drawn from the center perpendicular to a chord divides the chord into two equal parts.
MCQ 3. In a circle of radius 5 cm, a chord is 8 cm long. The distance of the chord from the center is:
- A. 3 cm
- B. 4 cm
- C. 5 cm
- D. 6 cm
Answer: A. 3 cm — By forming a right-angled triangle with the radius as hypotenuse, half the chord as one leg, and the distance from the center as the other leg, we can use the Pythagorean theorem to find the distance.
MCQ 4. The angle subtended by an arc at the center is 60 degrees. The angle subtended by the same arc at any point on the remaining part of the circle is:
- A. 30 degrees
- B. 60 degrees
- C. 90 degrees
- D. 120 degrees
Answer: A. 30 degrees — The angle subtended by an arc at the center is always double the angle subtended by the same arc at any point on the remaining part of the circle.
MCQ 5. In a circle, if ∠ACB = 40 degrees, where A, B, C are points on the circle, then ∠ADB (where D is another point on the circle on the same side of AB as C) is:
- A. 20 degrees
- B. 40 degrees
- C. 80 degrees
- D. 100 degrees
Answer: B. 40 degrees — Angles in the same segment of a circle are equal. Both ∠ACB and ∠ADB are subtended by the same arc AB.
MCQ 6. The angle in a semicircle is a:
- A. acute angle
- B. right angle
- C. obtuse angle
- D. reflex angle
Answer: B. right angle — The angle subtended by a diameter at any point on the circumference of a circle is always 90 degrees, which is a right angle.
MCQ 7. If ABCD is a cyclic quadrilateral and ∠A = 70 degrees, then ∠C is:
- A. 70 degrees
- B. 90 degrees
- C. 110 degrees
- D. 140 degrees
Answer: C. 110 degrees — In a cyclic quadrilateral, the sum of any pair of opposite angles is 180 degrees.
MCQ 8. In a cyclic quadrilateral ABCD, if ∠DBC = 80 degrees and ∠BAC = 40 degrees, then ∠BCD is:
- A. 60 degrees
- B. 70 degrees
- C. 80 degrees
- D. 100 degrees
Answer: A. 60 degrees — Using the property that angles in the same segment are equal (∠BDC = ∠BAC and ∠DAC = ∠DBC), we can find angles within triangle BCD and then find ∠BCD.
MCQ 9. Two parallel chords AB and CD of a circle are 7 cm apart. If the radius of the circle is 5 cm and the length of chord AB is 8 cm, find the length of chord CD.
- A. 6 cm
- B. 8 cm
- C. 9 cm
- D. 10 cm
Answer: A. 6 cm — First, find the distance of chord AB from the center using the Pythagorean theorem. Then, use the given distance between parallel chords to find the distance of chord CD from the center. Finally, use the Pythagorean theorem again to find the length of chord CD.
MCQ 10. If two arcs of a circle are congruent, then their corresponding chords are:
- A. unequal
- B. equal
- C. parallel
- D. perpendicular
Answer: B. equal — Congruent arcs subtend equal angles at the center, and this property directly leads to their corresponding chords being of equal length.
MCQ 11. The collection of all points in a plane, which are at a fixed distance from a fixed point in the plane, is called a:
- A. line
- B. square
- C. circle
- D. triangle
Answer: C. circle — This is the precise definition of a circle, where the fixed point is the center and the fixed distance is the radius.
MCQ 12. In a circle, O is the center. If ∠OAB = 30 degrees, then ∠ACB is: (A, B, C are points on the circle)
- A. 30 degrees
- B. 45 degrees
- C. 60 degrees
- D. 75 degrees
Answer: C. 60 degrees — Since OA and OB are radii, ΔOAB is an isosceles triangle, allowing us to find the central angle ∠AOB. The angle ∠ACB at the circumference is then half of ∠AOB.
MCQ 13. Two circles are congruent if they have the same:
- A. area
- B. circumference
- C. radius
- D. all of the above
Answer: D. all of the above — Congruent figures are identical in all respects. For circles, if their radii are equal, all other measurements like diameter, circumference, and area will also be equal, making them congruent.
MCQ 14. The locus of the centers of all circles passing through two given points is:
- A. the line joining the two points
- B. the perpendicular bisector of the segment joining the two points
- C. the circle passing through the two points
- D. the angle bisector of the angle formed by the two points with any other point
Answer: B. the perpendicular bisector of the segment joining the two points — For any circle passing through two points, its center must be equidistant from these two points. The locus of points equidistant from two fixed points is their perpendicular bisector.
MCQ 15. A chord of a circle is equal to its radius. The angle subtended by this chord at a point on the major arc is:
- A. 30 degrees
- B. 45 degrees
- C. 60 degrees
- D. 90 degrees
Answer: A. 30 degrees — If the chord is equal to the radius, the triangle formed by the chord and the two radii is equilateral, making the central angle 60 degrees. The angle on the major arc is then half of this central angle.
MCQ 16. ABCD is a cyclic quadrilateral. Side BC is extended to a point E. If ∠DCE = 110 degrees, then ∠BAD is:
- A. 70 degrees
- B. 90 degrees
- C. 110 degrees
- D. 140 degrees
Answer: C. 110 degrees — The exterior angle of a cyclic quadrilateral is equal to its interior opposite angle. Here, ∠DCE is the exterior angle, and ∠BAD is its interior opposite angle.
MCQ 17. A, B, C are three points on a circle such that the angle subtended by the chord AB at the center is 90 degrees and the angle subtended by the chord BC at the center is 110 degrees. Find the angle ∠ABC.
- A. 80 degrees
- B. 90 degrees
- C. 100 degrees
- D. 110 degrees
Answer: A. 80 degrees — By recognizing that OA=OB and OB=OC (radii), triangles OAB and OBC are isosceles. We can find ∠OBA and ∠OBC from the central angles and sum them to get ∠ABC.
MCQ 18. In a circle, if two chords AB and CD are equidistant from the center, then:
- A. AB > CD
- B. AB < CD
- C. AB = CD
- D. AB is perpendicular to CD
Answer: C. AB = CD — A key theorem in circles states that chords that are the same distance from the center of a circle are equal in length.
MCQ 19. If ABCD is a cyclic quadrilateral such that AB is a diameter of the circle and ∠ADC = 130 degrees, then ∠BAC is:
- A. 20 degrees
- B. 30 degrees
- C. 40 degrees
- D. 50 degrees
Answer: C. 40 degrees — Using the cyclic quadrilateral property, ∠ABC can be found. Since AB is a diameter, ∠ACB is 90 degrees. Then, use the angle sum property of ΔABC to find ∠BAC.
MCQ 20. How many circles can pass through three non-collinear points?
- A. One
- B. Two
- C. Three
- D. Infinitely many
Answer: A. One — Three non-collinear points define a unique triangle, and there is exactly one unique circle (the circumcircle) that passes through its three vertices.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.