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Chapter 10: CIRCLES

Grade 10 Mathematics · Chapter 10

Circles

Tangents, Secants, Theorems and Properties

◯ Tangent

⋅⋅ Secant

⊥ Radius-Tangent

≅ Equal Tangents

Intro & Definitions

Theorems & Proofs

10+

Worked Examples

A–D

Practice Sets

8pt

Exam Quick-Check

ⓘ	Introduction

A circle is the set of all points in a plane equidistant from a fixed point called the centre. In Class IX you explored basic circle terminology — chord, arc, sector, segment. In Chapter 10 we go further, examining the relationship between a circle and lines drawn in the same plane.

A line and a circle in a plane can interact in exactly three ways. Understanding these cases is the foundation of everything in this chapter.

Three Possible Positions of a Line and a Circle

Circle

← Line PQ (no intersection) →

Case (i): Non-Intersecting

The line PQ and the circle have no common point. PQ is called a non-intersecting line with respect to the circle.

A B

← Line PQ cuts at A and B →

Case (ii): Secant

The line PQ meets the circle at two distinct points A and B. PQ is called a secant of the circle. The chord AB is the part of the secant inside the circle.

← Line PQ touches at A only →

Case (iii): Tangent

The line PQ meets the circle at exactly one point A. PQ is called a tangent to the circle. Point A is called the point of contact.

📚	Key Definitions

Def. 1

Tangent to a Circle

A tangent to a circle is a line that intersects the circle at exactly one point. This unique point is called the point of contact. The tangent merely “touches” the circle — it does not cross into its interior. Every point on a circle has one and only one tangent at that point.

Def. 2

Secant

A secant is a line that intersects a circle at two distinct points. The portion of the secant inside the circle is a chord. A tangent can be seen as the limiting case of a secant when the two points of intersection gradually come together and coincide at one point.

Def. 3

Length of a Tangent

The length of the tangent from an external point P to a circle is the length of the segment from P to the point of contact T. It is the distance measured along the tangent line from the external point to where it touches the circle.

Def. 4

Point of Contact

The point of contact is the single point shared by the tangent line and the circle. The tangent is said to touch the circle at this point. The radius drawn to this point is always perpendicular to the tangent — this is the key geometric property of tangents.

#	Number of Tangents from a Point

• P (inside)

CASE 1: Point INSIDE

✘ No tangent possible. Every line through an interior point cuts the circle at two points — it is always a secant, never a tangent.

P ←(on circle)

← one tangent at P →

CASE 2: Point ON Circle

✓ Exactly one tangent. At any point on the circle, there is one and only one tangent. It is perpendicular to the radius at that point.

T₁ & T₂

• P
(outside)

CASE 3: Point OUTSIDE

✓✓ Exactly two tangents. From any external point P, exactly two tangents can be drawn to the circle. Their lengths are always equal: PT₁ = PT₂.

Δ	Theorems with Proofs

Theorem 10.1 The tangent at any point of a circle is perpendicular to the radius through the point of contact.

DIAGRAM

O (Centre)

O•
↓
r

Y ←—— P ——→ X

OP ⊥ XY (tangent)

Proof (by contradiction):

Given: Circle with centre O. Tangent XY touches the circle at point P. Radius = OP.

To prove: OP ⊥ XY

Step 1: Take any point Q on XY, other than P, and join OQ.

Step 2: Q must lie outside the circle (if Q were inside, XY would cut the circle at two points and be a secant, not a tangent — contradiction).

Step 3: Since Q is outside the circle, OQ > radius OP, i.e., OQ > OP.

Step 4: This holds for every point Q on XY except P. Therefore OP is the shortest distance from O to line XY.

Conclusion: The shortest distance from a point to a line is the perpendicular. Hence OP ⊥ XY. ■

Key Corollaries: (1) At any point on a circle, there exists one and only one tangent. (2) The line through the radius at the point of contact (perpendicular to tangent) is called the normal to the circle at that point.

Theorem 10.2 The lengths of the two tangents drawn from an external point to a circle are equal.

DIAGRAM

Q (contact)

O	• P (external)

R (contact)

PQ = PR (equal tangent lengths)

Proof (using RHS congruence):

Given: Circle with centre O, external point P, tangents PQ and PR with points of contact Q and R respectively.

To prove: PQ = PR

Construction: Join OP, OQ, and OR.

Step 1: By Theorem 10.1, ∠OQP = 90° and ∠ORP = 90° (radius ⊥ tangent).

Step 2: In right triangles OQP and ORP:

OQ = OR (radii of the same circle)

OP = OP (common side)

∠OQP = ∠ORP = 90°

Step 3: By RHS congruence: △OQP ≅ △ORP

Step 4: By CPCT: PQ = PR ■

Extra Insight: Also from CPCT, ∠OPQ = ∠OPR, so OP bisects the angle ∠QPR. This means the line from the external point to the centre bisects the angle between the two tangents.
Pythagoras alternative proof: PQ² = OP² − OQ² = OP² − OR² = PR², so PQ = PR.

✎	Worked Examples

▶ Example 1 — Chord of Larger Circle Touches Smaller Circle (Bisection Proof)

Problem:

In two concentric circles with centre O, a chord AB of the larger circle C₁ touches the smaller circle C₂ at point P. Prove that AP = BP (the chord is bisected at P).

Proof:

Step 1: AB is a tangent to the smaller circle C₂ at point P (since AB touches C₂ at exactly one point).

Step 2: Join OP. By Theorem 10.1, OP ⊥ AB (radius of C₂ is perpendicular to tangent AB at P).

Step 3: AB is a chord of the larger circle C₁, and OP ⊥ AB.

Step 4: By the theorem “the perpendicular from the centre to a chord bisects the chord,” OP bisects AB. Therefore AP = BP. ■

✓ Result: The chord of the larger circle that is tangent to the smaller circle is bisected at the point of tangency.

▶ Example 2 — Prove ∠PTQ = 2∠OPQ for Two Tangents from External Point

Problem:

Two tangents TP and TQ are drawn from external point T to a circle with centre O. Prove that ∠PTQ = 2∠OPQ.

Proof:

Let ∠PTQ = θ. By Theorem 10.2, TP = TQ, so △TPQ is isosceles.

Therefore: ∠TPQ = ∠TQP = ½(180° − θ) = 90° − θ/2

By Theorem 10.1: ∠OPT = 90° (radius ⊥ tangent at P)

∠OPQ = ∠OPT − ∠TPQ = 90° − (90° − θ/2) = θ/2 = ½∠PTQ

Therefore: ∠PTQ = 2∠OPQ ■

✓ The angle at the external point is twice the angle between the tangent and the line to centre.

▶ Example 3 — Find Length TP: Chord PQ = 8 cm, Radius = 5 cm (Tangents Meet at T)

Problem:

PQ is a chord of length 8 cm in a circle of radius 5 cm. Tangents at P and Q intersect at T. Find TP.

Solution:

OT bisects PQ perpendicularly at R (since △TPQ is isosceles with TP = TQ, and TO is the angle bisector).

So PR = RQ = 4 cm. OR = √(OP² − PR²) = √(25 − 16) = √9 = 3 cm.

In △TRP and △PRO: ∠TPR + ∠RPO = 90° = ∠TPR + ∠PTR → ∠RPO = ∠PTR

△TRP ~ △PRO (AA similarity): TP/PO = RP/RO → TP/5 = 4/3

Therefore TP = 20/3 cm.

✓ TP = 20/3 cm ≈ 6.67 cm

▶ Example 4 — Find Radius: Tangent Length 24 cm, Distance from Centre 25 cm

Problem:

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. Find the radius of the circle.

Solution:

Let the point of contact be T and the centre be O. By Theorem 10.1, OT ⊥ QT.

In right triangle OTQ: OQ² = OT² + QT²

25² = r² + 24² → 625 = r² + 576 → r² = 49 → r = 7 cm

✓ Radius = 7 cm [Answer: (A) 7 cm]

▶ Example 5 — Find ∠PTQ when ∠POQ = 110° (Two Tangents from T)

Problem:

TP and TQ are tangents to a circle with centre O, with ∠POQ = 110°. Find ∠PTQ.

Solution:

By Theorem 10.1: ∠OPT = 90° and ∠OQT = 90°.

In quadrilateral OPTQ: ∠OPT + ∠PTQ + ∠TQO + ∠QOP = 360°

90° + ∠PTQ + 90° + 110° = 360°

∠PTQ = 360° − 290° = 70°

✓ ∠PTQ = 70° [Answer: (B) 70°]

▶ Example 6 — Find ∠POA when Tangents PA and PB are at 80° to Each Other

Problem:

Tangents PA and PB from a point P to a circle with centre O are inclined to each other at 80°. Find ∠POA.

Solution:

∠APB = 80°. By Theorem 10.2, OP bisects ∠APB (since PA = PB, O lies on the angle bisector).

Therefore ∠OPA = 40°.

In right triangle OAP: ∠OAP = 90° (radius ⊥ tangent)

∠POA = 90° − ∠OPA = 90° − 40° = 50°

✓ ∠POA = 50° [Answer: (A) 50°]

▶ Example 7 — Find Radius: Tangent Length 4 cm from Point 5 cm from Centre

Problem:

A tangent from point A, which is 5 cm from the centre of a circle, has length 4 cm. Find the radius.

Solution:

Let the radius = r cm. By Pythagoras (radius ⊥ tangent): r² + 4² = 5²

r² = 25 − 16 = 9 → r = 3 cm

✓ Radius = 3 cm

▶ Example 8 — Chord of Larger Concentric Circle Touching Smaller (Chord Length)

Problem:

Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle that is tangent to the smaller circle.

Solution:

The chord AB touches the smaller circle at P. OP ⊥ AB where OP = 3 cm (radius of small circle).

OA = 5 cm (radius of large circle). By Pythagoras: AP = √(OA² − OP²) = √(25 − 9) = √16 = 4 cm.

Since OP bisects AB (perpendicular from centre): AB = 2 × AP = 2 × 4 = 8 cm.

✓ Length of chord = 8 cm

▶ Example 9 — Quadrilateral Circumscribing a Circle: AB + CD = AD + BC (Proof)

Problem:

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

ABCD circumscribes circle; P, Q, R, S are points of tangency on AB, BC, CD, DA

A	P	B
S		Q
D	R	C

Proof:

By Theorem 10.2 (tangents from an external point are equal):

From A: AP = AS | From B: BP = BQ | From C: CQ = CR | From D: DR = DS

Adding: AP + BP + CQ + DR = AS + BQ + CR + DS

(AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CR) → AB + CD = AD + BC ■

✓ AB + CD = AD + BC — the sum of opposite sides of a circumscribed quadrilateral are equal.

▶ Example 10 — Triangle ABC Circumscribes Circle of Radius 4 cm; Find AB and AC

Problem:

A triangle ABC circumscribes a circle of radius 4 cm. BC is divided by the point of contact D into BD = 8 cm and DC = 6 cm. Find AB and AC.

Solution:

Let tangent lengths from A = x, from B = y, from C = z.

From B: BD = 8 = y. From C: DC = 6 = z.

Semi-perimeter: s = (AB + BC + CA)/2 = (x+y + y+z + z+x)/2 = x + y + z = x + 8 + 6 = x + 14

Area of △ABC = r × s = 4(x + 14). Also Area = √[s(s−a)(s−b)(s−c)].

s − BC = x + 14 − 14 = x; s − AB = z = 6; s − AC = y = 8

Area = √[(x+14)·x·6·8] = 4(x+14) → (x+14)·48x = 16(x+14)²

48x = 16(x+14) → 48x = 16x + 224 → 32x = 224 → x = 7

AB = x + y = 7 + 8 = 15 cm | AC = x + z = 7 + 6 = 13 cm

✓ AB = 15 cm and AC = 13 cm

▶ Example 11 — Prove Tangents at Ends of a Diameter are Parallel

Problem:

Prove that tangents drawn at the two ends of a diameter of a circle are parallel to each other.

Proof:

←—— Tangent at A ——→

A •

diameter

• B

←—— Tangent at B ——→

Let AB be a diameter of circle with centre O. Let tangents at A and B be l₁ and l₂ respectively.

By Theorem 10.1: OA ⊥ l₁ → ∠OAl₁ = 90°

By Theorem 10.1: OB ⊥ l₂ → ∠OBl₂ = 90°

Since AB is a transversal cutting l₁ and l₂, and ∠OAl₁ = ∠OBl₂ = 90° (co-interior angles summing to 180°), l₁ ∥ l₂. ■

✓ Tangents at the ends of a diameter are always parallel.

✎	Practice Sets A – D

PRACTICE SET A — Tangent Basics and Pythagoras

A1. A tangent from point X to a circle has length 15 cm. The centre of the circle is 17 cm from X. Find the radius. [Answer: r² = 17² − 15² = 289 − 225 = 64 → r = 8 cm]

A2. From an external point P, the length of the tangent to a circle of radius 6 cm is 8 cm. Find the distance of P from the centre. [Answer: OP = √(6² + 8²) = √(36 + 64) = √100 = 10 cm]

A3. Fill in: A circle has ______ tangent(s) at any point on it; ______ tangent(s) from a point inside it; and ______ tangent(s) from an external point. [Answer: 1 ; 0 ; 2]

A4. A tangent PQ at P (circle radius 5 cm, centre O) meets a line through O at Q with OQ = 12 cm. Find PQ. [Answer: PQ = √(OQ² − OP²) = √(144 − 25) = √119 cm]

PRACTICE SET B — Angle Problems with Tangents

B1. Two tangents PA and PB from P to a circle with centre O make ∠APB = 60°. Find ∠AOB. [Answer: ∠OAP = ∠OBP = 90°; ∠AOB = 360° − 90° − 90° − 60° = 120°]

B2. Tangents TP and TQ to a circle have ∠PTQ = 50°. Find ∠OPQ (where O is the centre). [Answer: By Example 2, ∠OPQ = ½∠PTQ = 25°]

B3. In circle with centre O, tangents PA and PB are drawn. ∠POA = 40°. Find ∠APB. [Answer: ∠OAP = 90°; ∠OPA = 90° − 40° = 50°; by symmetry ∠APB = 2 × 50° = 100°]

B4. Tangents at A and B (ends of diameter) to a circle make angles α and β with a chord AC through A. Show the relationship between α and the arc AB. [Hint: Tangent-chord angle = inscribed angle in alternate segment; ∠ = half intercepted arc]

PRACTICE SET C — Concentric Circles and Chord Problems

C1. Two concentric circles have radii 13 cm and 5 cm. Find the length of the chord of the larger circle tangent to the smaller. [Answer: Half-chord = √(13² − 5²) = √(169 − 25) = √144 = 12 cm; full chord = 24 cm]

C2. In two concentric circles, the chord of the outer circle (radius 10) that is tangent to the inner circle has length 16 cm. Find the radius of the inner circle. [Answer: r² = 10² − 8² = 100 − 64 = 36 → r = 6 cm]

C3. A chord PQ of a circle of radius 10 cm has length 12 cm. Tangents at P and Q meet at T. Find TP. [Answer: PR = 6; OR = √(100−36) = 8; TP/PO = RP/RO → TP/10 = 6/8 → TP = 7.5 cm]

C4. A chord of a circle of radius r is tangent to a concentric circle of radius r/2. Find the length of the chord. [Answer: half-chord = √(r² − r²/4) = r√3/2; chord = r√3]

PRACTICE SET D — Mixed Proof and Application Questions

D1. Prove that the angle between two tangents from an external point is supplementary to the angle subtended at the centre by the chord joining the points of contact. [Hint: In quad OAPB, ∠OAP = ∠OBP = 90°; so ∠APB + ∠AOB = 180°]

D2. Prove that a parallelogram circumscribing a circle must be a rhombus. [Hint: By Q.8 result, AB + CD = AD + BC. In parallelogram AB = CD and AD = BC. So AB = BC, making all sides equal → rhombus]

D3. Prove that the perpendicular from the point of contact to the tangent passes through the centre. [Hint: By Theorem 10.1, the radius OP is perpendicular to tangent XY at P. Any perpendicular to XY at P must lie along OP, which passes through O]

D4. ABCD is a quadrilateral circumscribing a circle. AB = 12, BC = 10, CD = 9. Find AD. [Answer: AB + CD = AD + BC → 12 + 9 = AD + 10 → AD = 11 cm]

★	Chapter Summary

◯ Tangent Definition A tangent to a circle is a line touching it at exactly one point (point of contact). It is the limiting case of a secant when the two intersection points merge into one.	⊥ Theorem 10.1 The tangent at any point of a circle is perpendicular to the radius through the point of contact. So radius ⊥ tangent always.
≅ Theorem 10.2 The two tangents drawn from an external point to a circle are equal in length. Also OP bisects the angle between the two tangents.	▣ Circumscribed Polygon For a quadrilateral circumscribing a circle: AB + CD = AD + BC. For a triangle: sum of tangent pairs from each vertex using semi-perimeter method.
🔄 Number of Tangents 0 tangents from interior point · 1 tangent from point on circle · 2 tangents from exterior point. At most 2 tangents can be parallel to any given secant.	△ Key Angle Results ∠PTQ + ∠POQ = 180° (supplementary). ∠PTQ = 2∠OPQ. OP is the angle bisector of ∠QPR from external point P.

✅	8-Point Exam Quick-Check Master these before your exam — common errors and essential facts

① Radius ⊥ Tangent always Wherever a tangent touches a circle, the radius to that point is always perpendicular. This creates a right angle — the foundation of every tangent calculation.	② Equal tangents from external point From any external point, the two tangent lengths are always equal. Use this immediately when you see two tangents from the same external point — PT₁ = PT₂.
③ Angle sum in quadrilateral OAPB = 360° When two tangents from P touch at A and B, quadrilateral OAPB has ∠OAP = ∠OBP = 90°. So ∠APB + ∠AOB = 180° (supplementary). Never forget the two right angles.	④ Point inside circle → no tangent possible Any line through an interior point always cuts the circle in two points (secant). No tangent can be drawn from inside. This is a common MCQ trap.
⑤ Circumscribed quadrilateral: AB + CD = AD + BC Always use equal-tangent property from each vertex to prove this. Applies to any quadrilateral whose sides are all tangent to a circle (not just specific shapes).	⑥ Perpendicular from centre bisects chord Combined with Theorem 10.1, this helps solve concentric circle problems. If a chord of the outer circle is tangent to the inner circle, the perpendicular (= inner radius) bisects the chord.
⑦ Tangent-chord problems: use similar triangles For problems where tangents meet beyond a chord (like Example 3), look for similar triangles formed by the perpendicular bisector from the centre. This is the most efficient method.	⑧ Proof method: start with a diagram and label For every proof involving tangents, start by marking all right angles (radius ⊥ tangent), label equal lengths (equal tangents from same point), and identify congruent triangles (usually RHS or AA).

Grade 10 Maths Chapter 10 — Circles (NCERT) is a crucial chapter in CBSE Class 10 Mathematics, focusing on tangents to a circle, their properties, and related theorems. Students learn the definition of a tangent versus a secant, the concept of a point of contact, and the precise number of tangents possible from points inside, on, and outside the circle. The two core theorems — that the tangent is perpendicular to the radius at the point of contact, and that tangents from an external point are equal in length — form the foundation for all problem-solving in this chapter. Applications include concentric circle chord problems, angle relationships using quadrilaterals formed by two tangents and two radii, circumscribed polygons (particularly the AB + CD = AD + BC result for quadrilaterals), and proofs involving congruent triangles (RHS congruence). This comprehensive revision page includes full definitions, theorem proofs with diagrams, 11 worked examples, four graded practice sets with complete solutions, a chapter summary card, and an 8-point exam quick-check — ideal preparation for CBSE Class 10 board exams on circles and tangents.

Keywords: circles class 10, tangent to a circle, NCERT maths chapter 10 class 10, tangent perpendicular to radius, equal tangents from external point, CBSE grade 10 circles, tangent secant theorem

Curriculum

Mathematics Grade 10

Grade 10 Mathematics

Interactive Worksheets

Chapter Wise Worksheets

Solved Problems and solutions

Chapter 10: CIRCLES

Circles

Introduction

Key Definitions

Number of Tangents from a Point

Theorems with Proofs

Worked Examples

Practice Sets A – D

Chapter Summary

8-Point Exam Quick-Check