Circles — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Circles" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Circles" — 8 important questions with detailed answers for CBSE board exam prepara…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 10 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. State and prove the theorem: A tangent to a circle is perpendicular to the radius at the point of tangency.
Theorem: A tangent to a circle is perpendicular to the radius at the point of tangency.
Given:
Let O be the center of a circle.
Let AB be a tangent to the circle at point P.
OP is the radius to point P.
Prove: OP ⊥ AB (that is, OP is perpendicular to AB)
Proof:
Step 1: Assume, for contradiction, that OP is not perpendicular to AB.
Suppose OP makes an angle other than 90° with the tangent AB.
Step 2: From O, draw a perpendicular to the tangent AB, meeting it at point Q.
Since this perpendicular from O to line AB meets at Q, we have OQ ⊥ AB.
Step 3: Now, OQ is perpendicular to the tangent, so OQ is less than or equal to OP.
In fact, if Q ≠ P, then by the geometry of perpendiculars, OQ < OP.
Step 4: But we know OP is a radius to the point of contact P.
The definition of a tangent is a line that touches the circle at exactly one point and does not intersect the circle at any other point.
Step 5: If OQ < OP and Q is on the tangent line, then Q must be inside the circle (since Q is closer to O than P is).
But if Q is on the tangent line and inside the circle, this contradicts the fact that the tangent touches the circle at only one point.
Step 6: Therefore, our assumption must be false.
We must have Q = P, which means the perpendicular from O to the tangent at the point of contact coincides with OP.
Step 7: Conclusion.
OP ⊥ AB
Hence proved. ✓
Converse: If a line is perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle.
Q2. Two tangents are drawn to a circle from an external point. Prove that they are equal in length.
Theorem: Two tangents drawn to a circle from an external point are equal in length.
Given:
Let O be the center of the circle.
Let P be an external point.
PA and PB are two tangents to the circle at points A and B respectively.
Prove: PA = PB
Proof:
Step 1: Since PA and PB are tangents, we use the perpendicularity property.
OA ⊥ PA (radius is perpendicular to tangent at point of contact)
OB ⊥ PB (radius is perpendicular to tangent at point of contact)
Step 2: Consider triangles OAP and OBP.
In triangle OAP:
∠OAP = 90° (from Step 1)
In triangle OBP:
∠OBP = 90° (from Step 1)
Step 3: Identify equal sides.
OA = OB (both are radii of the same circle)
OP = OP (common side)
Step 4: Apply the RHS (Right Angle - Hypotenuse - Side) congruence criterion.
In triangles OAP and OBP:
∠OAP = ∠OBP = 90°
OP = OP (hypotenuse is common)
OA = OB (one side)
By RHS congruence: Triangle OAP ≅ Triangle OBP
Step 5: From congruent triangles, corresponding parts are equal.
PA = PB (corresponding sides of congruent triangles)
Hence proved. ✓
Additional Results:
1. The line joining P and O bisects the angle ∠APB.
2. The line PO bisects the chord AB at right angles.
Q3. A circle with center O and radius 5 cm has a point P at distance 13 cm from the center. From P, two tangents are drawn to the circle. Find the length of each tangent.
Given:
Circle with center O and radius r = 5 cm
Point P at distance OP = 13 cm from center
Two tangents PA and PB are drawn from P to the circle
Find: Length of each tangent (PA and PB)
Diagram (text description):
A (point of tangency)
/|
/ |
/ | 5 cm (radius OA)
/ |
/ |
/ O________
/ 13 cm
/
P \_____ B (point of tangency)
Solution:
Step 1: Since PA is a tangent at point A:
OA ⊥ PA
Therefore, triangle OAP is a right triangle with right angle at A.
Step 2: In right triangle OAP, apply the Pythagorean theorem.
OP² = OA² + PA²
13² = 5² + PA²
169 = 25 + PA²
PA² = 169 - 25
PA² = 144
PA = 12 cm
Step 3: Similarly, for tangent PB at point B:
OB ⊥ PB
Triangle OBP is right-angled at B.
Step 4: In right triangle OBP:
OP² = OB² + PB²
13² = 5² + PB²
169 = 25 + PB²
PB² = 144
PB = 12 cm
Alternatively, by the theorem proved earlier:
PA = PB (tangents from an external point are equal)
So PB = 12 cm
Answer: Length of each tangent = 12 cm
Verification:
13² = 5² + 12²
169 = 25 + 144
169 = 169 ✓ (Pythagorean triple: 5-12-13)
Q4. A chord of a circle is 24 cm long. The radius of the circle is 13 cm. Find the distance of the chord from the center of the circle.
Given:
Length of chord = 24 cm
Radius of circle = 13 cm
Find: Distance of chord from the center
Diagram (text description):
O (center)
|
| d (perpendicular distance)
|
----M---- (M is foot of perpendicular, midpoint of chord)
/ | \
A | B
|--12-|-12--| (half-chord on each side)
\__________/
Chord AB = 24 cm
Key Property:
The perpendicular from the center of a circle to a chord bisects the chord.
Step 1: Let O be the center and AB be the chord with length 24 cm.
Let M be the foot of the perpendicular from O to AB.
M is the midpoint of AB.
Step 2: Find AM (half the chord).
AM = AB / 2 = 24 / 2 = 12 cm
Step 3: In right triangle OMA:
∠OMA = 90° (perpendicular)
OA = 13 cm (radius)
AM = 12 cm
Step 4: Apply the Pythagorean theorem.
OA² = OM² + AM²
13² = OM² + 12²
169 = OM² + 144
OM² = 169 - 144
OM² = 25
OM = 5 cm
Answer: Distance of chord from the center = 5 cm
Verification:
13² = 5² + 12²
169 = 25 + 144
169 = 169 ✓ (Pythagorean triple: 5-12-13)
Q5. Two circles with centers O₁ and O₂ and radii 8 cm and 5 cm respectively touch each other externally. Find the distance between their centers.
Given:
Circle 1: Center O₁, radius r₁ = 8 cm
Circle 2: Center O₂, radius r₂ = 5 cm
The circles touch each other externally
Find: Distance between centers O₁O₂
Diagram (text description):
Circle 1 Point of tangency Circle 2
* * *
\ / \ /
\ 8 cm / \ 5 cm /
\ O₁ O₂ /
*-----------•---------•-----------*
8 cm ? 5 cm
Definition:
Two circles touch externally when they have exactly one point in common, and one circle lies outside the other.
Step 1: Understand external tangency.
When two circles touch externally:
- The point of tangency lies on the line joining the centers
- The distance between centers = sum of the radii
Step 2: Apply the formula for external tangency.
Distance between centers = r₁ + r₂
O₁O₂ = 8 + 5
O₁O₂ = 13 cm
Answer: Distance between centers = 13 cm
Note:
For internal tangency (one circle inside the other, touching at one point):
Distance between centers = |r₁ - r₂| = |8 - 5| = 3 cm
Q6. A circle has a radius of 10 cm. A point P is at distance 26 cm from the center. From P, a secant is drawn that passes through the center O. This secant intersects the circle at points A and B (where A is closer to P). Find PA and PB.
Given:
Circle with center O and radius r = 10 cm
Point P at distance OP = 26 cm from center
Secant through P and O intersects circle at A and B
A is closer to P than B
Find: PA and PB
Diagram (text description):
A O B
•----•---------•----P
10 cm 10 cm 16 cm
Secant line through O
Solution:
Step 1: Understand the configuration.
Since the secant passes through the center O, it's actually a diameter extended to point P.
Let's place O at the origin with the secant line horizontal.
Step 2: Identify positions on the secant.
On the line POB (from left to right):
- A is on the circle, on the side closer to P
- O is the center
- B is on the circle, on the opposite side
- P is external
Actually, let's reconsider. If OP = 26 and the radius is 10:
- A is at distance OA = 10 on one side
- B is at distance OB = 10 on the opposite side
- P is at distance OP = 26 from O
Step 3: Since the secant passes through O:
PA = OP - OA = 26 - 10 = 16 cm
(This is the distance from P to the near intersection point)
PB = OP + OB = 26 + 10 = 36 cm
(This is the distance from P to the far intersection point)
Answer:
PA = 16 cm
PB = 36 cm
Verification using Power of a Point:
Power of point P = PA × PB = 16 × 36 = 576
Also, Power of point P = OP² - r² = 26² - 10² = 676 - 100 = 576 ✓
Q7. In a circle with radius 6 cm, a chord subtends an angle of 60° at the center. Find the length of the chord.
Given:
Radius of circle = 6 cm
Angle subtended by chord at center = 60°
Find: Length of chord
Diagram (text description):
O (center)
/|
/ | 6 cm (radius)
/ |
6 cm | 60° angle
/ |
A-----B (chord)
Solution:
Step 1: Consider the triangle OAB formed by the two radii and the chord.
OA = OB = 6 cm (radii)
∠AOB = 60° (angle at center)
Step 2: Since OA = OB, triangle OAB is isosceles.
Step 3: Find the base angles.
In isosceles triangle OAB:
∠OAB = ∠OBA = (180° - 60°) / 2 = 120° / 2 = 60°
Step 4: Since all angles are 60°, the triangle is equilateral.
Step 5: In an equilateral triangle, all sides are equal.
AB = OA = OB = 6 cm
Answer: Length of chord = 6 cm
Alternative Method using Chord Length Formula:
For a chord subtending angle θ at the center of a circle with radius r:
Chord length = 2r sin(θ/2)
Chord length = 2 × 6 × sin(60°/2)
= 12 × sin(30°)
= 12 × (1/2)
= 6 cm ✓
Q8. Two chords AB and CD of a circle intersect at point P inside the circle. If AP = 4 cm, PB = 6 cm, and CP = 3 cm, find PD.
Given:
Two chords AB and CD intersect at point P inside a circle
AP = 4 cm, PB = 6 cm, CP = 3 cm
Find: PD
Diagram (text description):
C
|
/ | \
/ | \
/ | \
A-----P-----B
\ | /
\ | /
\ | /
D
Theorem (Intersecting Chords):
If two chords of a circle intersect at a point P inside the circle, then:
AP × PB = CP × PD
This is known as the Power of a Point theorem for intersecting chords.
Step 1: Apply the intersecting chords theorem.
AP × PB = CP × PD
4 × 6 = 3 × PD
24 = 3 × PD
Step 2: Solve for PD.
PD = 24 / 3
PD = 8 cm
Answer: PD = 8 cm
Verification:
AP × PB = 4 × 6 = 24
CP × PD = 3 × 8 = 24 ✓
Note:
This theorem applies to any two chords intersecting inside a circle, regardless of whether they're perpendicular or at any other angle.
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 that intersects a circle at exactly one point is called a:
- A. Chord
- B. Secant
- C. Tangent
- D. Radius
Answer: C. Tangent — By definition, a tangent is a line that touches the circle at only one point, called the point of contact.
MCQ 2. How many tangents can be drawn to a circle from a point lying inside the circle?
- A. One
- B. Two
- C. Zero
- D. Infinitely many
Answer: C. Zero — A line passing through a point inside the circle will always intersect the circle at two points or not at all, thus it cannot be a tangent.
MCQ 3. The tangent at any point of a circle is __________ to the radius through the point of contact.
- A. Parallel
- B. Perpendicular
- C. Equal
- D. Intersecting
Answer: B. Perpendicular — This is a fundamental theorem of circles, stating that the radius drawn to the point of tangency is always perpendicular to the tangent.
MCQ 4. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
- A. 13 cm
- B. 8.5 cm
- C. √119 cm
- D. 7 cm
Answer: C. √119 cm — In right-angled ΔOPQ (∠OPQ = 90°), by Pythagoras theorem, OQ² = OP² + PQ². So, 12² = 5² + PQ², which gives PQ = √119 cm.
MCQ 5. How many tangents can be drawn to a circle from a point lying on the circle?
- A. One
- B. Two
- C. Zero
- D. Infinitely many
Answer: A. One — From a point on the circle, only one unique line can be drawn that touches the circle at that specific point, which is the tangent.
MCQ 6. How many tangents can be drawn to a circle from a point lying outside the circle?
- A. One
- B. Two
- C. Zero
- D. Three
Answer: B. Two — From any external point, exactly two tangents can be drawn to a given circle. These tangents have equal lengths from the external point to the points of contact.
MCQ 7. If the lengths of tangents drawn from an external point P to a circle are 8 cm and (3x + 2) cm, what is the value of x?
- A. 2
- B. 6
- C. 8/3
- D. 10/3
Answer: A. 2 — The lengths of tangents drawn from an external point to a circle are equal. So, 3x + 2 = 8, which implies 3x = 6, and thus x = 2.
MCQ 8. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
- A. 8 cm
- B. 10 cm
- C. 12 cm
- D. 16 cm
Answer: A. 8 cm — Let R=5 cm, r=3 cm. The radius to the point of tangency (r) is perpendicular to the chord. Using Pythagoras, the half-chord length is √(R² - r²) = √(5² - 3²) = √16 = 4 cm. So the full chord length is 2 * 4 = 8 cm.
MCQ 9. If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to:
- A. 60°
- B. 70°
- C. 80°
- D. 90°
Answer: B. 70° — In quadrilateral OPTQ, ∠OPT = ∠OQT = 90° (radius ⊥ tangent). The sum of angles is 360°, so ∠PTQ = 360° - (90° + 90° + 110°) = 360° - 290° = 70°.
MCQ 10. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
- A. 7 cm
- B. 12 cm
- C. 10 cm
- D. 8 cm
Answer: A. 7 cm — Let O be the centre and P the point of contact. ΔOPQ is a right-angled triangle with ∠OPQ = 90°. By Pythagoras theorem, OQ² = OP² + PQ². So, 25² = OP² + 24², which yields OP = 7 cm.
MCQ 11. A line that intersects a circle at two distinct points is called a:
- A. Chord
- B. Tangent
- C. Secant
- D. Arc
Answer: C. Secant — A secant is a line that intersects a circle at two distinct points. A chord is a line segment connecting two points on the circle, which is part of a secant.
MCQ 12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the length of side AB.
- A. 12 cm
- B. 13 cm
- C. 14 cm
- D. 15 cm
Answer: D. 15 cm — Using tangent properties, let AE = AF = x. Then AB = x+8, AC = x+6, BC = 14. Equating Area = rs with Heron's formula leads to x=7. Thus, AB = 7 + 8 = 15 cm.
MCQ 13. In a circle, two parallel tangents are drawn. The distance between them is 14 cm. The radius of the circle is:
- A. 7 cm
- B. 14 cm
- C. 3.5 cm
- D. 28 cm
Answer: A. 7 cm — Parallel tangents are drawn at the ends of a diameter. The distance between them is equal to the diameter. So, diameter = 14 cm, and radius = 14/2 = 7 cm.
MCQ 14. A quadrilateral ABCD is drawn to circumscribe a circle. If AB=6 cm, BC=7 cm, CD=4 cm, then AD is:
- A. 3 cm
- B. 4 cm
- C. 5 cm
- D. 6 cm
Answer: A. 3 cm — For a quadrilateral circumscribing a circle, the sum of opposite sides is equal. So, AB + CD = BC + AD. 6 + 4 = 7 + AD, which gives AD = 3 cm.
MCQ 15. Two circles touch externally at a point P. A common tangent AB to the circles is drawn, where A and B are points on the circles. If the radius of the first circle is 9 cm and the radius of the second circle is 4 cm, what is the length of the common tangent AB?
- A. 12 cm
- B. 13 cm
- C. 10 cm
- D. 15 cm
Answer: A. 12 cm — Let radii be r1=9 and r2=4. The distance between centers is r1+r2=13. Drawing a line through the smaller center parallel to AB forms a right triangle with hypotenuse 13 and one leg (r1-r2)=5. The other leg, which is AB, is √(13²-5²) = √144 = 12 cm.
MCQ 16. A circle touches all four sides of a quadrilateral ABCD. If AB = 7 cm, BC = 8 cm, CD = 5 cm, then the perimeter of the quadrilateral is:
- A. 20 cm
- B. 22 cm
- C. 24 cm
- D. 26 cm
Answer: C. 24 cm — For a quadrilateral circumscribing a circle, AB + CD = BC + AD. So, 7 + 5 = 8 + AD, which gives AD = 4 cm. Perimeter = 7 + 8 + 5 + 4 = 24 cm.
MCQ 17. The common point of a tangent to a circle and the circle is called the:
- A. Centre
- B. Endpoint
- C. Point of contact
- D. Intersection point
Answer: C. Point of contact — The unique point where a tangent line touches a circle is known as the point of contact or point of tangency.
MCQ 18. From a point P, 10 cm away from the center of a circle, the length of the tangent PT is 8 cm. The radius of the circle is:
- A. 4 cm
- B. 5 cm
- C. 6 cm
- D. 7 cm
Answer: C. 6 cm — In right-angled ΔOTP (∠T = 90°), OP is the hypotenuse = 10 cm, PT = 8 cm. By Pythagoras theorem, OT² = OP² - PT² = 10² - 8² = 100 - 64 = 36. So, OT = 6 cm.
MCQ 19. If a chord AB of a circle with centre O subtends an angle of 80° at the centre, then the angle between the tangents at A and B is:
- A. 100°
- B. 80°
- C. 50°
- D. 40°
Answer: A. 100° — Let tangents meet at P. In quadrilateral OAPB, ∠OAP = ∠OBP = 90° (radius ⊥ tangent). Sum of angles in a quadrilateral is 360°, so ∠APB = 360° - (90° + 90° + 80°) = 100°.
MCQ 20. A circle can have __________ parallel tangents at the most.
- A. One
- B. Two
- C. Three
- D. Four
Answer: B. Two — A circle can have at most two parallel tangents, which are always drawn at the ends of a diameter.
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.