Class 10Maths · GeometryFull chapter

Circles

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Tangent to a CircleTangent is Perpendicular to the RadiusNumber of Tangents from a PointEqual Tangents from an External PointSolving Problems with the Two Theorems

Tangent to a Circle

Quick answer A tangent is a straight line that touches a circle at exactly one point (the point of contact); a secant is a line that cuts the circle at two points.

Take a circle and a straight line drawn in the same plane. Exactly three situations are possible:

  • The line does not meet the circle at all (a non-intersecting line).
  • The line meets the circle at two points — this is called a secant.
  • The line meets the circle at exactly one point — this is called a tangent.

The single common point of the tangent and the circle is called the point of contact. The word tangent comes from the Latin tangere, meaning “to touch”.

A tangent can be viewed as the limiting case of a secant. Imagine a secant PQ that cuts a circle at two points P and Q. If we slowly turn the line so that Q slides towards P, the two points come nearer and nearer. When they finally merge into a single point, the secant has become a tangent. So a tangent is the position of a secant when its two points of intersection coincide.

Worked example: How many tangents can be drawn to a circle at a given point that lies on the circle? At any point on a circle there is one and only one tangent. So exactly one tangent passes through a point lying on the circle. (A circle as a whole, however, has infinitely many tangents — one at each of its infinitely many points.)

Points of contact of a tangent number of common points = 1 A tangent meets the circle at exactly one point.
Tangents at a point on the circle exactly 1 tangent Only one tangent can be drawn at a point lying on the circle.
Remember
  • A tangent touches a circle at exactly one point; a secant cuts it at two points.
  • The common point of a tangent and a circle is the point of contact.
  • A tangent is the limiting position of a secant whose two intersection points coincide.
  • There is exactly one tangent at any given point on a circle.
  • A circle has infinitely many tangents in all (one at every point of the circle).

Tangent is Perpendicular to the Radius

Quick answer Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius drawn to the point of contact, so the radius meets the tangent at 90°.

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

Idea of proof: Let a tangent touch a circle with centre O at point P. Take any other point Q on the tangent line (other than P). Since the tangent meets the circle only at P, every other point Q of the tangent lies outside the circle. Therefore OQ is longer than the radius OP for every such Q. This means OP is the shortest distance from O to the tangent line. The shortest distance from a point to a line is the perpendicular distance. Hence OP is perpendicular to the tangent, i.e. OP ⊥ tangent, so the angle between the radius and the tangent at P is 90°.

This single fact makes a right angle available in almost every circle problem, letting us use the Pythagoras theorem in the right triangle formed by the radius, the tangent segment, and the line joining the centre to the external point.

Worked example: A tangent to a circle with centre O touches it at point P. A point Q lies on this tangent with OQ = 25 cm, and the radius OP = 7 cm. Find the length PQ.

  1. By Theorem 10.1, OP ⊥ PQ, so triangle OPQ is right-angled at P.
  2. By the Pythagoras theorem, OQ² = OP² + PQ².
  3. 25² = 7² + PQ² ⇒ 625 = 49 + PQ² ⇒ PQ² = 576.
  4. PQ = √576 = 24 cm.
Tangent–radius angle (Theorem 10.1) OP ⊥ tangent, angle = 90° O is the centre, P is the point of contact.
Right triangle relation at point of contact OQ² = OP² + PQ² Pythagoras in right triangle OPQ, right angle at P; Q on the tangent.
Remember
  • The tangent at any point is perpendicular to the radius through the point of contact (angle = 90°).
  • Every point of the tangent except the point of contact lies outside the circle.
  • The radius to the point of contact is the shortest distance from the centre to the tangent line.
  • This 90° angle lets you apply the Pythagoras theorem in circle problems.
  • Converse: a line through the end of a radius, perpendicular to it, is a tangent.

Number of Tangents from a Point

Quick answer From a point inside a circle no tangent can be drawn, from a point on the circle exactly one, and from a point outside the circle exactly two tangents can be drawn.

The number of tangents that can be drawn to a circle from a given point depends on where the point lies relative to the circle.

  • Point inside the circle: No tangent can be drawn. Any line through an interior point is a secant — it cuts the circle at two points, so it can never touch at only one point.
  • Point on the circle: Exactly one tangent can be drawn, namely the tangent at that point.
  • Point outside the circle: Exactly two tangents can be drawn. These are the two tangent lines that touch the circle from that external point.

For an external point P, the two tangent segments (from P to the two points of contact) are called the lengths of the tangents from P. The line joining the external point to the centre passes exactly between the two tangents and bisects the angle between them.

Worked example: A point P lies at a distance of 10 cm from the centre O of a circle of radius 6 cm. How many tangents can be drawn from P, and what is the length of each tangent?

  1. Since OP = 10 cm > radius 6 cm, the point P lies outside the circle, so two tangents can be drawn.
  2. Let the tangent touch the circle at A. Then OA ⊥ PA (Theorem 10.1), so triangle OAP is right-angled at A.
  3. PA² = OP² − OA² = 10² − 6² = 100 − 36 = 64.
  4. PA = √64 = 8 cm. Each tangent from P has length 8 cm.
Number of tangents by position inside → 0, on → 1, outside → 2 Depends on the position of the point relative to the circle.
Length of tangent from an external point ℓ = √(d² − r²) cm · d = distance of the external point from the centre, r = radius.
Remember
  • From a point inside the circle: zero tangents can be drawn.
  • From a point on the circle: exactly one tangent can be drawn.
  • From a point outside the circle: exactly two tangents can be drawn.
  • The two tangents from an external point have equal length.
  • The line from the external point to the centre bisects the angle between the two tangents.

Equal Tangents from an External Point

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

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

Proof: Let two tangents from an external point P touch a circle with centre O at points A and B. Join OA, OB and OP.

  1. OA ⊥ PA and OB ⊥ PB (tangent is perpendicular to the radius, Theorem 10.1). So ∠OAP = ∠OBP = 90°.
  2. In right triangles OAP and OBP: OA = OB (radii of the same circle), and OP = OP (common hypotenuse).
  3. By the RHS congruence rule, △OAP ≅ △OBP.
  4. Hence by CPCT, PA = PB. The two tangent lengths are equal.

As a bonus from the same congruent triangles, ∠OPA = ∠OPB, so OP bisects ∠APB, and ∠AOP = ∠BOP, so OP bisects ∠AOB as well.

Worked example: From an external point P, two tangents PA and PB are drawn to a circle. If PA = 12 cm, find PB. Also, if ∠APB = 70°, find ∠APO.

  1. By Theorem 10.2, PB = PA = 12 cm.
  2. OP bisects ∠APB, so ∠APO = ½ × 70° = 35°.
Equal tangents (Theorem 10.2) PA = PB PA, PB are tangents from external point P touching at A, B.
Angle bisected by centre line ∠OPA = ∠OPB = ½∠APB OP bisects the angle between the two tangents.
Remember
  • The two tangents drawn from an external point to a circle are equal in length (PA = PB).
  • The proof uses RHS congruence of the two right triangles formed with the centre.
  • The line joining the external point to the centre bisects the angle between the tangents.
  • It also bisects the angle subtended by the two radii at the centre.
  • This equal-tangents result is used constantly in problems on triangles and quadrilaterals circumscribing a circle.

Solving Problems with the Two Theorems

Quick answer Circle problems are solved by combining the 90° radius–tangent angle (Pythagoras) with the equal-tangents property, especially for triangles and quadrilaterals drawn around a circle.

Almost every board problem on this chapter is built from just two facts: the tangent is perpendicular to the radius (Theorem 10.1) and equal tangents from an external point (Theorem 10.2). A useful derived result for a quadrilateral drawn around a circle follows directly from equal tangents.

Key result (quadrilateral circumscribing a circle): If a quadrilateral ABCD is drawn to circumscribe a circle (all four sides are tangents), then the sums of opposite sides are equal:

AB + CD = AD + BC

This is because the two tangents from each vertex are equal: from A they are equal, from B equal, from C equal and from D equal; adding side by side gives the result.

Worked example 1: A quadrilateral ABCD circumscribes a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, find AD.

  1. By the key result, AB + CD = AD + BC.
  2. 6 + 4 = AD + 7 ⇒ 10 = AD + 7 ⇒ AD = 3 cm.

Worked example 2: Two concentric circles have radii 5 cm and 3 cm. Find the length of a chord of the larger circle that touches (is tangent to) the smaller circle.

  1. Let the chord AB of the bigger circle touch the smaller circle at M. Then OM is a radius of the small circle drawn to the point of contact, so OM ⊥ AB and OM = 3 cm.
  2. The perpendicular from the centre bisects the chord, so AM = MB and OA = 5 cm (radius of large circle).
  3. In right triangle OMA: AM² = OA² − OM² = 5² − 3² = 25 − 9 = 16, so AM = 4 cm.
  4. Chord AB = 2 × AM = 2 × 4 = 8 cm.
Quadrilateral circumscribing a circle AB + CD = AD + BC Sum of one pair of opposite sides equals sum of the other pair.
Chord tangent to inner concentric circle chord = 2√(R² − r²) cm · R = radius of larger circle, r = radius of smaller circle.
Remember
  • Draw the radius to every point of contact — it is perpendicular to the tangent (creates a 90° angle for Pythagoras).
  • Mark equal tangent segments from each external point to set up equations.
  • For a quadrilateral circumscribing a circle: AB + CD = AD + BC.
  • A perpendicular from the centre to a chord bisects the chord (used with concentric circles).
  • If a circle touches all sides of a parallelogram, that parallelogram must be a rhombus.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

number of common points = 1
Points of contact of a tangent
exactly 1 tangent
Tangents at a point on the circle
OP ⊥ tangent, angle = 90°
Tangent–radius angle (Theorem 10.1)
OQ² = OP² + PQ²
Right triangle relation at point of contact
inside → 0, on → 1, outside → 2
Number of tangents by position
ℓ = √(d² − r²)
Length of tangent from an external pointcm
PA = PB
Equal tangents (Theorem 10.2)
∠OPA = ∠OPB = ½∠APB
Angle bisected by centre line
AB + CD = AD + BC
Quadrilateral circumscribing a circle
chord = 2√(R² − r²)
Chord tangent to inner concentric circlecm

Test yourself

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0 correct · 0/12 answered
Q1 Tangent basics easy

A straight line that touches a circle at exactly one point is called a:

Q2 Number of tangents easy

How many tangents can be drawn to a circle from a point lying outside it?

Q3 Tangent perpendicular to radius easy

The tangent at any point of a circle is ______ to the radius through the point of contact.

Q4 Length of tangent medium

A point P is 13 cm from the centre of a circle of radius 5 cm. The length of the tangent from P is:

Q5 Equal tangents easy

From an external point P, two tangents PA and PB are drawn to a circle. If PA = 10 cm, then PB equals:

Q6 Concentric circles medium

Two concentric circles have radii 13 cm and 5 cm. The length of the chord of the larger circle that touches the smaller circle is:

Q7 Angle between tangents medium

Two tangents drawn from an external point are inclined to each other at 70°. The angle subtended by the two radii at the centre is:

Q8 Circumscribing quadrilateral medium

A quadrilateral ABCD circumscribes a circle with AB = 6 cm, BC = 7 cm and CD = 4 cm. Then AD equals:

Q9 Tangent perpendicular to radius easy

The angle between a tangent to a circle and the radius drawn to the point of contact is:

Q10 Tangents and angles hard

Tangents PA and PB are drawn from external point P to a circle of radius 6 cm. If ∠APB = 60°, the distance OP is:

Q11 Length of tangent medium

A tangent PQ touches a circle of radius 8 cm at Q, and OP = 17 cm where O is the centre. The length PQ is:

Q12 Angle between tangents hard

From a point at distance 2r from the centre of a circle of radius r, the two tangents are inclined to each other at an angle of:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 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.Length of tangent

Let O be the centre and PQ the tangent touching the circle at P, so OP is the radius.

  1. The tangent is perpendicular to the radius at the point of contact, so OP ⊥ PQ and triangle OPQ is right-angled at P.
  2. By the Pythagoras theorem: OQ² = OP² + PQ².
  3. 25² = OP² + 24² ⇒ 625 = OP² + 576.
  4. OP² = 625 − 576 = 49, so OP = √49 = 7 cm.

The radius of the circle is 7 cm.

2 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.Concentric circles

Let O be the common centre. Let AB be a chord of the larger circle (radius 5 cm) that touches the smaller circle (radius 3 cm) at point M.

  1. OM is the radius of the smaller circle drawn to the point of contact, so OM ⊥ AB and OM = 3 cm.
  2. The perpendicular from the centre to a chord bisects it, so AM = MB.
  3. In right triangle OMA (right angle at M), OA = 5 cm (radius of larger circle): AM² = OA² − OM² = 5² − 3² = 25 − 9 = 16.
  4. AM = √16 = 4 cm, so AB = 2 × AM = 2 × 4 = 8 cm.

The length of the chord is 8 cm.

3 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 find ∠POA.Angle between tangents

PA and PB are tangents from external point P, touching the circle at A and B, with ∠APB = 80°.

  1. OA ⊥ PA, so ∠OAP = 90°.
  2. OP bisects ∠APB (line to the centre bisects the angle between the tangents), so ∠OPA = ½ × 80° = 40°.
  3. In triangle OPA, the angles sum to 180°: ∠POA = 180° − ∠OAP − ∠OPA = 180° − 90° − 40°.
  4. ∠POA = 50°.

Therefore ∠POA = 50°.

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

Let the circle touch the sides AB, BC, CD and DA at the points P, Q, R and S respectively.

  1. Tangents drawn from an external point are equal in length. Applying this at each vertex: AP = AS (from A), BP = BQ (from B), CR = CQ (from C), DR = DS (from D).
  2. Add these four equations: AP + BP + CR + DR = AS + BQ + CQ + DS.
  3. Group the tangents that make up each side: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ).
  4. That is AB + CD = AD + BC.

Hence AB + CD = AD + BC, proved.

5 In a triangle ABC, a circle is inscribed touching side BC at D such that BD = 8 cm and DC = 6 cm. If the radius of the circle is 4 cm, find the sides AB and AC.Incircle of a triangle

Let the incircle touch BC, CA and AB at D, E and F respectively, with centre O and radius 4 cm. Let AF = AE = x.

  1. Equal tangents from each vertex: BF = BD = 8 cm, CE = CD = 6 cm, AF = AE = x.
  2. So AB = x + 8, AC = x + 6 and BC = 8 + 6 = 14 cm.
  3. Semi-perimeter s = (AB + BC + CA)/2 = ((x+8) + 14 + (x+6))/2 = x + 14.
  4. Area = r × s = 4(x + 14). Also, by Heron's formula, Area = √[s(s−a)(s−b)(s−c)] where s−BC = x, s−AB = 6, s−AC = 8, so Area = √[(x+14)(x)(6)(8)] = √[48x(x+14)].
  5. Equate: 4(x+14) = √[48x(x+14)]. Squaring: 16(x+14)² = 48x(x+14).
  6. Divide by 16(x+14): (x + 14) = 3x ⇒ 2x = 14 ⇒ x = 7.

Therefore AB = 7 + 8 = 15 cm and AC = 7 + 6 = 13 cm.

6 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.Tangents at ends of diameter

Let AB be a diameter of a circle with centre O. Let PQ be the tangent at A and RS be the tangent at B.

  1. The tangent at a point is perpendicular to the radius (and hence the diameter) through that point.
  2. At A: AB ⊥ PQ, so ∠PAB = 90°.
  3. At B: AB ⊥ RS, so ∠RBA = 90°.
  4. Now AB is a transversal cutting the two lines PQ and RS. The alternate interior angles ∠PAB and ∠RBA are each 90°, hence equal.
  5. Since a pair of alternate interior angles are equal, the two lines are parallel.

Therefore PQ ∥ RS: the tangents at the ends of a diameter are parallel, proved.

Previous-year board questions 4

Q1 Prove that the lengths of tangents drawn from an external point to a circle are equal. CBSE 2023 3 marks

Let PA and PB be two tangents drawn from an external point P to a circle with centre O, touching it at A and B. Join OA, OB and OP.

  1. Since a tangent is perpendicular to the radius at the point of contact, OA ⊥ PA and OB ⊥ PB, so ∠OAP = ∠OBP = 90°.
  2. In right triangles OAP and OBP: OA = OB (radii of the same circle), OP = OP (common), and ∠OAP = ∠OBP = 90°.
  3. By the RHS congruence rule, △OAP ≅ △OBP.
  4. By CPCT, PA = PB.

Hence the lengths of the two tangents from an external point are equal. (Proved)

Q2 Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. CBSE 2019 3 marks

Let a tangent XY touch a circle with centre O at point P. We prove OP ⊥ XY.

  1. Take any point Q on the tangent XY other than P, and join OQ.
  2. Since XY is a tangent, it meets the circle only at P; every other point of XY, including Q, lies outside the circle.
  3. Therefore OQ is greater than the radius OP, i.e. OQ > OP, for every position of Q other than P.
  4. So OP is the shortest of all distances from O to the line XY.
  5. The shortest distance from a point to a line is the perpendicular distance. Hence OP ⊥ XY.

Therefore the tangent at any point of a circle is perpendicular to the radius through the point of contact. (Proved)

Q3 The length of a tangent drawn from a point A to a circle is 4 cm and the distance of A from the centre O is 5 cm. Find the radius of the circle. CBSE 2020 1 mark

Let the tangent from A touch the circle at P, so OP is the radius.

  1. OP ⊥ AP (tangent perpendicular to radius), so triangle OPA is right-angled at P.
  2. By the Pythagoras theorem: OA² = OP² + AP².
  3. 5² = OP² + 4² ⇒ 25 = OP² + 16 ⇒ OP² = 9.
  4. OP = √9 = 3 cm.

The radius of the circle is 3 cm.

Q4 Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. CBSE 2018 3 marks

Let ∠PTQ = θ. TP and TQ touch the circle at P and Q.

  1. Tangents from an external point are equal, so TP = TQ, making triangle TPQ isosceles.
  2. In triangle TPQ, the base angles are equal: ∠TPQ = ∠TQP = (180° − θ) / 2 = 90° − θ/2.
  3. OP is a radius and TP is a tangent, so OP ⊥ TP, giving ∠OPT = 90°.
  4. Now ∠OPQ = ∠OPT − ∠TPQ = 90° − (90° − θ/2) = θ/2.
  5. So ∠OPQ = θ/2 = ½ ∠PTQ.

Therefore ∠PTQ = 2∠OPQ. (Proved)

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