⭕ Circles

NCERT Class 10 Mathematics | Chapter 10 | Exercises 10.1 & 10.2 | Interactive Solutions

🎯 Key Concepts from Chapter 10

These theorems and properties form the foundation of all circle problems in this chapter.

📐 Tangent Definition

  • A line that intersects a circle at exactly one point.
  • At the point of contact, the tangent is perpendicular to the radius.

📏 Secant Definition

  • A line that intersects a circle at two distinct points.
  • Every secant contains a chord of the circle.

⚡ Theorem 10.1

  • The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  • This creates right triangles in almost every tangent problem!

⚡ Theorem 10.2

  • The lengths of tangents drawn from an external point to a circle are equal.
  • If PA and PB are tangents from P, then PA = PB.

🔺 Right Triangle Strategy

  • Always look for the right angle where radius meets tangent.
  • Use Pythagoras theorem: (hypotenuse)² = (radius)² + (tangent)²

📐 Quadrilateral Properties

  • For quadrilateral circumscribing circle: AB + CD = AD + BC
  • Parallelogram circumscribing circle is always a rhombus.
📚 Essential Formulas & Theorems
Theorem 10.1: Tangent ⊥ Radius at point of contact
If PT is tangent at T, then OT ⊥ PT, so ∠OTP = 90°
Theorem 10.2: Tangents from external point are equal
If PA and PB are tangents from P, then PA = PB
Pythagoras in Tangent Problems:
(Distance from external point to centre)² = (Radius)² + (Tangent length)²
Circumscribed Quadrilateral:
AB + CD = AD + BC (sum of opposite sides equal)
Opposite sides subtend supplementary angles at centre.
💡 Problem-Solving Strategy
1 Identify the tangent — Mark the point of contact and draw the radius to it.
2 Mark the right angle — Radius ⊥ Tangent always gives 90°.
3 Use Pythagoras — For numerical problems, apply a² + b² = c².
4 Equal tangents — From same external point, tangents are equal (Theorem 10.2).
5 Angle sums — In quadrilaterals, sum = 360°. Use this for angle problems.

📘 Exercise 10.1

4 Questions | Short Answer & Fill in the Blanks

1
Short Answer
How many tangents can a circle have?
O Infinitely many tangents possible One tangent at every point on circumference
Fig. 10.1 — A circle has infinitely many tangents, one at each point on its circumference
✅ Step-by-Step Solution
1A circle consists of infinitely many points on its circumference.
2At each and every point on the circumference, exactly one tangent can be drawn.
3Since there are infinitely many points, there are infinitely many tangents.
🎯 Final Answer: Infinite (uncountably many)
2
Fill in the Blanks
Fill in the blanks:
O Secant A B Tangent Tangent: 1 point | Secant: 2 points
Fig. 10.2 — Tangent touches at one point, Secant cuts at two points

(i) A tangent to a circle intersects it in point(s).

(ii) A line intersecting a circle in two points is called a .

(iii) A circle can have parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called .

3
MCQ
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:
O r=5 Tangent PQ OQ=12 P Q ⊾ Right angle at P (radius ⊥ tangent)
Fig. 10.3 — Right triangle OPQ where OP ⊥ PQ (radius ⊥ tangent)
✅ Step-by-Step Solution
1Since PQ is tangent at P, OP ⊥ PQ (radius is perpendicular to tangent at point of contact).
2Therefore, ΔOPQ is a right triangle with right angle at P.
3Given: OP = radius = 5 cm, OQ = 12 cm
4By Pythagoras theorem: OQ² = OP² + PQ²
5PQ² = OQ² - OP² = 12² - 5² = 144 - 25 = 119
6PQ = √119 cm
🎯 Final Answer: √119 cm
4
Construction
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
O Tangent (m) Secant (n) Given line (l) m || n || l | m at distance = radius
Fig. 10.4 — Line m (tangent) at distance = radius, Line n (secant) at distance < radius
✅ Construction Steps
1Draw a circle with centre O and a given line l.
2Draw a line m parallel to l at distance equal to the radius from centre O → this line touches the circle at exactly one point, so it is a tangent.
3Draw another line n parallel to l at distance less than the radius from centre O → this line cuts the circle at two points, so it is a secant.
4The tangent line is at exactly radius distance from centre; the secant is closer to the centre than the radius.
🎯 Result: Line m = Tangent, Line n = Secant, both parallel to given line l

📗 Exercise 10.2

13 Questions | MCQs, Proofs & Numericals

1
MCQ
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:
O P Q r Tangent d (distance) ⊾ Right ΔOPQ: d² = r² + t²
Fig. 10.5 — Right triangle formed by radius, tangent, and line to external point
✅ Step-by-Step Solution
1Let O be the centre, P be the point of contact, and Q be the external point.
2Given: PQ = 24 cm (tangent length), OQ = 25 cm
3Since tangent ⊥ radius at point of contact: OP ⊥ PQ
4In right ΔOPQ: OQ² = OP² + PQ² (Pythagoras theorem)
5OP² = OQ² - PQ² = 25² - 24² = 625 - 576 = 49
6OP = √49 = 7 cm
🎯 Final Answer: 7 cm
2
MCQ
In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to:
O P Q T r r Tangent Tangent ∠PTQ = ? ∠POQ = 110° (given)
Fig. 10.11 — Two tangents from external point T, forming quadrilateral OPTQ
✅ Step-by-Step Solution
1Given: TP and TQ are tangents from external point T.
2Since radius ⊥ tangent: ∠OPT = 90° and ∠OQT = 90°
3In quadrilateral OPTQ, sum of angles = 360°
4∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360°
5110° + 90° + ∠PTQ + 90° = 360°
6∠PTQ = 360° - 290° = 70°
🎯 Final Answer: 70°
3
MCQ
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to:
O A B P OP bisects ∠AOB ∠APB = 80° Find ∠POA
Fig. 10.11 variant — OP bisects ∠AOB due to congruent triangles
✅ Step-by-Step Solution
1Given: PA and PB are tangents from P, ∠APB = 80°
2Since radius ⊥ tangent: ∠OAP = 90° and ∠OBP = 90°
3In quadrilateral OAPB: ∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
4∠AOB + 90° + 80° + 90° = 360°
5∠AOB = 360° - 260° = 100°
6Since tangents from external point are equal (PA = PB), and OA = OB (radii),
7ΔOAP ≅ ΔOBP (RHS congruence), so OP bisects ∠AOB
8∠POA = ∠AOB/2 = 100°/2 = 50°
🎯 Final Answer: 50°
4
Proof
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
O A B Diameter AB l₁ (Tangent at A) l₂ (Tangent at B) ⊾ ⊾ Both ⊥ to AB ⇒ l₁ || l₂
Fig. 10.6 — Tangents at ends of diameter are both perpendicular to the diameter, hence parallel
✅ Proof
1Let AB be a diameter of circle with centre O.
2Let l₁ be tangent at A and l₂ be tangent at B.
3Since radius ⊥ tangent: OA ⊥ l₁ and OB ⊥ l₂
4Therefore, ∠OAl₁ = 90° and ∠OBl₂ = 90°
5Since AOB is a straight line (diameter), OA and OB are collinear.
6Thus l₁ and l₂ are both perpendicular to the same line AB.
7Lines perpendicular to the same line are parallel.
8Hence, l₁ || l₂. Proved.
🎯 Hence Proved: Tangents at ends of diameter are parallel
5
Proof
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
O P Tangent l Radius OP ⊾ Perpendicular at P This perpendicular passes through O
Fig. 10.7 — The perpendicular at P to tangent l must pass through centre O (uniqueness of perpendicular)
✅ Proof
1Let line l be tangent to circle at point P.
2We know that radius OP ⊥ tangent l (Theorem 10.1), so ∠OPl = 90°.
3At point P on line l, there is exactly ONE perpendicular line (uniqueness of perpendicular).
4Since OP is perpendicular to l at P, the perpendicular at P to l must be the line OP itself.
5Therefore, the perpendicular at P to tangent l passes through the centre O.
🎯 Hence Proved: Perpendicular at point of contact passes through centre
6
MCQ
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
O P Q r Tangent d (distance) ⊾ Right ΔOPQ: d² = r² + t²
Fig. 10.8 — Same right triangle setup: radius² = distance² - tangent²
✅ Step-by-Step Solution
1Let O be the centre, P be the point of contact, and A be the external point.
2Given: OA = 5 cm, AP = 4 cm (tangent length)
3Since OP ⊥ AP (radius ⊥ tangent): ΔOPA is right-angled at P
4By Pythagoras: OA² = OP² + AP²
5OP² = OA² - AP² = 5² - 4² = 25 - 16 = 9
6OP = √9 = 3 cm
🎯 Final Answer: 3 cm
7
MCQ
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.
O Chord AB A B P OP=3 R=5 OP ⊥ AB, AP = PB, AB = 2×AP
Fig. 10.9 — Chord AB of larger circle touches smaller circle at P, forming right triangle OPA
✅ Step-by-Step Solution
1Let O be the common centre. Larger circle radius = 5 cm, smaller = 3 cm.
2Let AB be chord of larger circle touching smaller circle at P.
3Since tangent ⊥ radius at point of contact: OP ⊥ AB
4Also, perpendicular from centre to chord bisects the chord: AP = PB
5In right ΔOPA: OA² = OP² + AP²
6AP² = OA² - OP² = 5² - 3² = 25 - 9 = 16
7AP = 4 cm
8Therefore, AB = 2 × AP = 8 cm
🎯 Final Answer: 8 cm
8
Proof
A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC.
O A B C D P Q R S AP=AS, BP=BQ, CQ=CR, DR=DS
Fig. 10.12 — Tangents from same external point to a circle are equal in length
✅ Proof
1Let the circle touch AB at P, BC at Q, CD at R, and DA at S.
2From external point, tangents to a circle are equal in length:
3AP = AS (tangents from A) ... (i)
4BP = BQ (tangents from B) ... (ii)
5CR = CQ (tangents from C) ... (iii)
6DR = DS (tangents from D) ... (iv)
7LHS: AB + CD = (AP + PB) + (CR + RD)
8= AS + BQ + CQ + DS [using (i), (ii), (iii), (iv)]
9= (AS + DS) + (BQ + CQ) = AD + BC = RHS
🎯 Hence Proved: AB + CD = AD + BC
9
Proof
In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.
O XY X'Y' AB C OA OB Prove ∠AOB = 90°
Fig. 10.13 — Tangent AB intersects two parallel tangents XY and X'Y'
✅ Proof
1Join OC. Since AB is tangent at C: OC ⊥ AB.
2Join OA and OB. Let OA intersect the circle at P and OB at Q.
3In ΔOPA and ΔOCA: OP = OC (radii), OA = OA (common), ∠OPA = ∠OCA = 90°
4So ΔOPA ≅ ΔOCA (RHS), giving ∠POA = ∠COA ... (i)
5Similarly, ΔOQB ≅ ΔOCB (RHS), giving ∠QOB = ∠COB ... (ii)
6Since XY || X'Y' and both are tangents, P-O-Q is a diameter (straight line).
7∠POA + ∠COA + ∠COB + ∠QOB = 180° (straight angle)
82∠COA + 2∠COB = 180° [from (i) and (ii)]
9∠COA + ∠COB = 90°
10Therefore, ∠AOB = 90°. Proved.
🎯 Hence Proved: ∠AOB = 90°
10
Proof
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
O A B P ∠APB + ∠AOB = 180° Supplementary angles
Fig. 10.10 — ∠APB + ∠AOB = 180° (supplementary angles)
✅ Proof
1Let P be external point, PA and PB be tangents, A and B points of contact.
2Given to prove: ∠APB + ∠AOB = 180°
3In quadrilateral OAPB: ∠OAP = 90° and ∠OBP = 90° (radius ⊥ tangent)
4Sum of angles in quadrilateral = 360°
5∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
6∠AOB + 90° + ∠APB + 90° = 360°
7∠AOB + ∠APB = 360° - 180° = 180°
8Hence, ∠APB and ∠AOB are supplementary. Proved.
🎯 Hence Proved: ∠APB + ∠AOB = 180°
11
Proof
Prove that the parallelogram circumscribing a circle is a rhombus.
O A B C D ABCD is parallelogram + circumscribed = Rhombus
Fig. 10.11 — Parallelogram ABCD circumscribing a circle must be a rhombus
✅ Proof
1Let ABCD be a parallelogram circumscribing a circle.
2From Q8 property: AB + CD = AD + BC (for any circumscribed quadrilateral)
3Since ABCD is parallelogram: AB = CD and AD = BC (opposite sides equal)
4Substituting: AB + AB = AD + AD
52AB = 2AD
6AB = AD
7Since adjacent sides AB = AD, and AB = CD, AD = BC, all four sides are equal.
8Therefore, ABCD is a rhombus. Proved.
🎯 Hence Proved: Parallelogram circumscribing a circle is a rhombus
12
MCQ
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 (see Fig. 10.14). Find the sides AB and AC.
O A B C D r=4 BD=8 DC=6 Find AB and AC using r × s = Area
Fig. 10.14 — Using equal tangents property and Area = r × s formula
✅ Step-by-Step Solution
1Let the circle touch AB at F, BC at D, and CA at E.
2Tangents from same external point are equal:
3BD = BF = 8 cm, CD = CE = 6 cm
4Let AF = AE = x cm
5Then AB = AF + FB = x + 8, AC = AE + EC = x + 6, BC = BD + DC = 14 cm
6Semi-perimeter s = (AB + BC + CA)/2 = (x+8 + 14 + x+6)/2 = x + 14
7Area of ΔABC = r × s = 4 × (x + 14)
8Also by Heron's formula: Area² = s(s-AB)(s-BC)(s-CA)
9= (x+14)(x+14-x-8)(x+14-14)(x+14-x-6) = (x+14)(6)(x)(8) = 48x(x+14)
10So [4(x+14)]² = 48x(x+14)
1116(x+14)² = 48x(x+14)
1216(x+14) = 48x
13x + 14 = 3x ⇒ 2x = 14 ⇒ x = 7 cm
14AB = x + 8 = 15 cm, AC = x + 6 = 13 cm
🎯 Final Answer: AB = 15 cm and AC = 13 cm
13
Proof
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
O A B C D α β γ δ ∠AOB + ∠COD = (α+β)+(γ+δ) = 180°
Fig. 10.15 — Opposite sides AB and CD subtend ∠AOB and ∠COD which are supplementary
✅ Proof
1Let ABCD be quadrilateral circumscribing circle with centre O.
2Let circle touch AB at P, BC at Q, CD at R, DA at S.
3Join OP, OQ, OR, OS. These are radii to points of contact.
4In ΔOAP and ΔOAS: OA = OA, OP = OS (radii), AP = AS (tangents from A)
5So ΔOAP ≅ ΔOAS (SSS), giving ∠AOP = ∠AOS = α (say)
6Similarly: ∠BOP = ∠BOQ = β, ∠COQ = ∠COR = γ, ∠DOR = ∠DOS = δ
7At centre: 2α + 2β + 2γ + 2δ = 360° ⇒ α + β + γ + δ = 180°
8Angle subtended by AB at centre = ∠AOB = α + β
9Angle subtended by CD at centre = ∠COD = γ + δ
10∠AOB + ∠COD = (α + β) + (γ + δ) = 180°
11Similarly, ∠BOC + ∠DOA = 180°
12Hence, opposite sides subtend supplementary angles at centre. Proved.
🎯 Hence Proved: Opposite sides subtend supplementary angles at centre

🧮 Circle Calculators

Interactive tools for tangent, secant, and circumscribed figure problems

📐 Tangent Length Calculator

Find tangent length given radius and distance from centre

Tangent Length: 12.00 cm
Valid: Distance > Radius ✓
⭕ Radius from Tangent

Find radius given tangent length and distance from centre

Radius: 7.00 cm
📏 Chord Length Calculator

For concentric circles (Q7 type)

Chord Length: 8.00 cm
🔺 Circumscribed Triangle Solver

Find sides AB and AC (Q12 type)

AB: 15.00 cm
AC: 13.00 cm
📐 Two Tangents Angle Calculator

Find angle between tangents (Q2/Q3 type)

Angle between tangents ∠PTQ: 70°
Half angle ∠POA: 55°
📊 Quadrilateral Property Checker

Verify AB + CD = AD + BC for circumscribed quadrilateral

AB + CD = 12, AD + BC = 12 ✓ Valid

📚 NCERT Class 10 Mathematics | Chapter 10: Circles

Exercises 10.1 & 10.2 | 17 Questions with Interactive Solutions & SVG Figures