Circles Class 10: Notes, Tangent Theorems and Questions

This guide covers Circles Class 10, chapter 10 of NCERT Class 10 Mathematics, which is about tangents to a circle. You will learn the two key theorems with their proofs, how many tangents can be drawn from a point, and how to use these ideas for chords, angles and quadrilaterals around a circle. Important questions with answers and common mistakes close the guide.

A line and a circle: three cases

A circle is the set of all points in a plane at a fixed distance (the radius) from a fixed point (the centre).

A line and a circle in the same plane can meet in three ways:

• The part of a secant inside the circle is a chord.
• The single common point of a tangent and the circle is the point of contact.
• The word tangent comes from the Latin word tangere, meaning to touch.

A tangent as a limiting case of a secant

Take a secant meeting the circle at P and Q. Keep P fixed and turn the line so that Q slides along the circle towards P. When Q reaches P, the secant becomes the tangent at P.

Another view: draw lines parallel to a secant. The chords get shorter as the lines move towards the edge. At the edge the chord length becomes zero and the line is a tangent. So a given secant has exactly two tangents parallel to it, one on each side. Their points of contact are the ends of a diameter perpendicular to the secant.

Theorem 10.1: tangent is perpendicular to the radius

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

Proof: let XY be the tangent at P to a circle with centre O. Take any point Q on XY other than P. Since XY touches the circle only at P, Q lies outside the circle, so OQ > OP. This is true for every point of XY except P. So OP is the shortest distance from O to XY, and the shortest distance from a point to a line is the perpendicular. Hence OP ⊥ XY.

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

Example: a tangent at P meets a line through the centre O at Q, with radius 5 cm and OQ = 12 cm. Since OP ⊥ PQ, OQ² = OP² + PQ², so 144 = 25 + PQ², PQ² = 119 and PQ = √119 cm.

How many tangents from a point?

The length of a tangent from an external point means the segment from that point to the point of contact. The full tangent line goes on in both directions.

Theorem 10.2: tangents from an external point are equal

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

Proof: let PQ and PR be tangents from P, touching at Q and R. Join O to P, Q and R.

1. ∠OQP = ∠ORP = 90° (Theorem 10.1).
2. OQ = OR (radii) and OP is common.
3. So ΔOQP ≅ ΔORP by RHS congruence.
4. Hence PQ = PR by CPCT.

From the same congruence, ∠OPQ = ∠OPR. So OP bisects the angle between the two tangents; the centre lies on that angle bisector.

Concentric circles, angles and quadrilaterals

Concentric circles share a centre. If a chord of the larger circle touches the smaller circle, the point of contact bisects the chord. The reason: the small circle's radius is perpendicular to the chord there, and a perpendicular from the centre to a chord bisects it.

Example: radii 5 cm and 3 cm. In right ΔOPA, AP = √(5² − 3²) = √16 = 4 cm, so the chord AB = 8 cm.

Angle between tangents: with tangents TP and TQ from T, the quadrilateral OPTQ has ∠OPT = ∠OQT = 90°. The angles of a quadrilateral add to 360°, so

∠PTQ + ∠POQ = 180°

If ∠POQ = 110°, then ∠PTQ = 70°. It is also true that ∠PTQ = 2∠OPQ.

Quadrilateral around a circle: if ABCD circumscribes a circle touching it at P, Q, R, S, then AP = AS, BP = BQ, CR = CQ and DR = DS. Adding and regrouping:

AB + CD = AD + BC

A parallelogram around a circle has AB = CD and BC = AD, so 2AB = 2BC and AB = BC. It must be a rhombus.

Remember this

• Tangent: exactly one common point with the circle.
• Tangent ⊥ radius at the point of contact.
• Inside point: 0 tangents. On the circle: 1. Outside: 2, of equal length.
• The centre lies on the bisector of the angle between two tangents.
• ∠PTQ + ∠POQ = 180°.
• Circumscribed quadrilateral: AB + CD = AD + BC.

Important questions with answers

1. How many common points does a tangent have with a circle? Exactly one. With two, it would be a secant.

2. How many tangents can be parallel to a given secant? Exactly two, one on each side of the circle.

3. A tangent PQ at P meets a line through the centre O at Q. The radius is 5 cm and OQ = 12 cm. Find PQ. OQ² = OP² + PQ², so PQ² = 144 − 25 = 119 and PQ = √119 cm.

4. A point is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from the point. The tangent is perpendicular to the radius, so length² = 13² − 5² = 169 − 25 = 144. Length = 12 cm.

5. One tangent from P to a circle is 8 cm long. How long is the other? Also 8 cm, by Theorem 10.2.

6. Two concentric circles have radii 5 cm and 3 cm. Find the chord of the larger circle that touches the smaller one. Half-chord = √(25 − 9) = 4 cm, so the chord is 8 cm.

7. The angle between two tangents from a point is 80°. Find the angle at the centre between the radii to the points of contact. 180° − 80° = 100°.

8. ABCD circumscribes a circle with AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD. AB + CD = AD + BC, so 10 = AD + 7 and AD = 3 cm.

9. Prove that a parallelogram circumscribing a circle is a rhombus. For a circumscribed quadrilateral, AB + CD = AD + BC. In a parallelogram AB = CD and AD = BC, so 2AB = 2BC, giving AB = BC. A parallelogram with equal adjacent sides is a rhombus.

Common mistakes to avoid

• Drawing a tangent from a point inside the circle. None exists.
• Forgetting to state why an angle is 90° (radius ⊥ tangent) in proofs.
• Writing RHS as the congruence reason without naming the right angle, hypotenuse and side.
• Adding ∠PTQ and ∠POQ to 360° instead of 180°.
• Pairing the wrong sides in AB + CD = AD + BC. Use opposite sides.

Practise both theorem proofs until you can write them from memory, and study this chapter with Joy to try more tangent problems.