Circles – Class 10 Maths Chapter 10 Notes & NCERT PDF

Class 10MathematicsChapter 10NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 10 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 10: Circles

Introduction

In Class 9 we learnt that a chord is a segment joining two points on a circle. This chapter studies the different ways a line and a circle can meet and focuses on tangents — lines that touch the circle at exactly one point.

Key Concepts & Theorems

Line and circle — three situations

Non-intersecting line: no common point with the circle.

Secant: a line that meets the circle at two points.

Tangent: a line that meets the circle at exactly one point; that point is the point of contact.

The word tangent comes from the Latin tangere, "to touch"; it was introduced by the Danish mathematician Thomas Fineke in 1583.

Facts about tangents:

A tangent is a special case of a secant when the two end points of its chord coincide.

There is only one tangent at a point of a circle.

The common point of the tangent and the circle is called the point of contact.

Theorem 10.1

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

Proof idea: every point Q on the tangent other than the point of contact P lies outside the circle, so OQ > OP. Hence OP is the shortest distance from O to the tangent, and the shortest segment from a point to a line is the perpendicular. So OP ⟂ tangent.

The line through the point of contact containing the radius is called the normal to the circle at that point.

Number of tangents from a point

Point inside the circle: no tangent.

Point on the circle: exactly one tangent.

Point outside the circle: exactly two tangents.

The length of the segment of the tangent from the external point to the point of contact is called the length of the tangent.

Theorem 10.2

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

Proof: in right triangles OQP and OQR (right angles at the points of contact), OQ = OQ and OP = OR (radii), so they are congruent by RHS; hence PQ = RQ. It also follows that ∠OQP = ∠OQR — the centre lies on the bisector of the angle between the two tangents.

Methods Step by Step

Join the centre to the point(s) of contact — the radius makes 90° with the tangent.

Look for right triangles and use Pythagoras: (length of tangent)² = (distance from centre)² – r².

Use equal tangents from the same external point to split side lengths (very useful in quadrilaterals or triangles circumscribing a circle).

In the two-tangent figure, angle between tangents + angle at centre = 180°.

Worked Examples

Concentric circles: a chord of the larger circle that touches the smaller circle is bisected at the point of contact (OP ⟂ AB, and a perpendicular from the centre bisects a chord).

Two tangents TP and TQ from T: ∠PTQ = 2∠OPQ (using TP = TQ in an isosceles triangle and ∠OPT = 90°).

PQ is a chord of length 8 cm of a circle of radius 5 cm; tangents at P and Q meet at T. Find TP. OT bisects PQ at R, so PR = 4 cm and OR = √(25 – 16) = 3 cm. Let TR = y. From △TRP and △TPO: TP² = y² + 16 = (y + 3)² – 25 ⇒ y = 16/3. So TP² = 256/9 + 16 = 400/9 ⇒ TP = 20/3 cm.

Tangent length from a point 5 cm from the centre of a circle of radius 3 cm: √(25 – 9) = 4 cm.

Point 13 cm from the centre, radius 5 cm: tangent length = √(169 – 25) = 12 cm.

Quadrilateral circumscribing a circle: using equal tangents at each vertex, AB + CD = AD + BC.

Board Exam Focus

Proofs of Theorem 10.1 and Theorem 10.2 are frequently asked (3 marks).

Short questions on the number of tangents from a point, and tangent length by Pythagoras.

"Prove that the parallelogram circumscribing a circle is a rhombus" and "AB + CD = AD + BC" type proofs.

Angle questions: the angle between two tangents from an external point is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Formula Sheet

Radius ⟂ tangent at point of contact

Tangent length from P: PT = √(OP² – r²)

Tangents from an external point are equal: TP = TQ

∠PTQ + ∠POQ = 180°

Circumscribed quadrilateral: AB + CD = AD + BC

Common Mistakes

Saying a point inside the circle has one tangent — it has none.

Forgetting to join the centre to the point of contact, so the right angle is never used.

Using the diameter instead of the radius in Pythagoras.

Assuming the two tangents from a point make a right angle with each other — they make right angles only with the radii.

In the RHS proof, not stating which parts are equal.

💡 Key Learning Points

✓Distinguish a non-intersecting line, a secant and a tangent to a circle.
✓Prove and use Theorem 10.1: the tangent at any point is perpendicular to the radius through the point of contact.
✓Know that a point inside, on or outside a circle has 0, 1 or 2 tangents respectively.
✓Prove and use Theorem 10.2: tangents drawn from an external point are equal in length.
✓Solve problems using right triangles formed by the radius, tangent and the line to the centre.

👨‍🏫 Teaching Tips

→Slide a ruler across a drawn circle to show the change from non-intersecting line to secant to tangent.
→Use a pin (centre), thread and a straight edge to physically show that the radius to the point of contact is the shortest distance to the tangent.
→Ask students to draw a point inside, on and outside a circle and try to draw tangents — let them discover 0, 1, 2.
→Emphasise 'join the centre to the point of contact' as the first move in almost every problem.
→Work the TP = 20/3 cm example slowly, showing how two right triangles give two equations for TP².
→Practise the circumscribed-quadrilateral result AB + CD = AD + BC with coloured equal tangent segments.

📋 Assessment Questions

Can the student write the proof that the tangent is perpendicular to the radius?
Can the student prove that tangents from an external point are equal (RHS congruence)?
Can the student find the length of a tangent using Pythagoras?
Can the student state how many tangents can be drawn from points inside, on and outside a circle?
Can the student prove AB + CD = AD + BC for a quadrilateral circumscribing a circle?

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