Tangents and Intersecting Chords

Q: A point is 26 cm from the centre of a circle of radius 10 cm. Find the length of the tangent from the point.

√(262 - 102) = √(676 - 100) = √(576) = 24 cm.

Q: PA and PB are tangents to a circle with centre O, and ∠ AOB = 110°. Find ∠ APB and ∠ PAB.

∠ APB = 180° - 110° = 70°. Triangle PAB is isosceles, so ∠ PAB = (180° - 70°) ÷ 2 = 55°.

ICSE Class 10 · Mathematics · Chapter 18

What this chapter covers

Tangents, secants and the point of contact
The tangent is perpendicular to the radius
Touching circles and the line of centres
Two tangents from an external point
Chords intersecting inside a circle
Chords intersecting outside a circle
A chord and a tangent meeting outside the circle
The alternate segment theorem
Using the theorems together
Common mistakes and how to avoid them

Key points

A tangent meets a circle at exactly one point, the point of contact; a secant cuts it at two points.
The tangent at any point is perpendicular to the radius through that point; the tangent length from P is √(OP² - r²).
When two circles touch, the point of contact lies on the line of centres; the distance between centres is r₁ + r₂ for external contact and r₁ - r₂ for internal contact.
Two tangents from an external point are equal; the line from the point to the centre bisects the angle between them, and ∠ APB + ∠ AOB = 180^°.
A quadrilateral circumscribing a circle has AB + CD = AD + BC; the tangent length from a vertex of a triangle to its incircle is the semi-perimeter minus the opposite side.
Chords intersecting internally or externally at P: PA × PB = PC × PD, every length measured from P.
A secant PAB and a tangent PT from the same external point: PA × PB = PT².
Alternate segment theorem: the angle between a tangent and a chord through the point of contact equals the angle in the alternate segment.
Proofs of all six theorems are part of the syllabus and should be learnt with their reasons.

Questions answered

Define a tangent to a circle and its point of contact.

A tangent to a circle is a straight line that meets the circle at exactly one point; that point is called the point of contact.

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

Every point Q of the tangent other than the point of contact P lies outside the circle, so OQ > OP. OP is therefore the shortest segment from O to the line, and the shortest segment from a point to a line is the perpendicular, so OP ⊥ tangent.

Why must the point of contact of two touching circles lie on the line of centres?

The two radii drawn to the point of contact are both perpendicular to the common tangent at that point. Only one perpendicular can be drawn to a line at a given point, so the two radii lie along one straight line through both centres.

Two circles of radii 9 cm and 5 cm touch internally. How far apart are their centres?

9 - 5 = 4 cm.

A point is 26 cm from the centre of a circle of radius 10 cm. Find the length of the tangent from the point.

√(26² - 10²) = √(676 - 100) = √(576) = 24 cm.

PA and PB are tangents to a circle with centre O, and ∠ AOB = 110^°. Find ∠ APB and ∠ PAB.

∠ APB = 180^° - 110^° = 70^°. Triangle PAB is isosceles, so ∠ PAB = (180^° - 70^°) ÷ 2 = 55^°.

The incircle of a triangle with sides 7 cm, 8 cm and 9 cm touches the sides. Find the tangent length from the vertex opposite the 9 cm side.

The semi-perimeter is 12 cm, so the tangent length is 12 - 9 = 3 cm.

Chords AB and CD meet at P inside a circle. PA = 9 cm, PB = 4 cm and PD = 3 cm. Find PC.

9 × 4 = PC × 3, so PC = 12 cm.

Past board exam questions

ICSE 2027 sample paper · 1 mark

The radius of the circle with centre O is 3 cm. PA and PB are the tangents to the circle which are at right angle to each other. The length of OP is:

(3)/(√(2)) cm
3 cm
3√(2) cm
6√(2) cm

Show answer

3√(2) cm

ICSE 2027 sample paper · 3 marks

The parallelogram ABCD circumscribes a circle, touching the circle at P, Q, R and S (P on AB, Q on BC, R on CD and S on DA). Prove that: (a) AB = BC (b) diagonals AC and BD bisect each other at right angles.

Show answer

(a) Tangents drawn from an external point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS. Adding: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), i.e. AB + CD = AD + BC. In a parallelogram AB = CD and AD = BC, so 2AB = 2BC, i.e. AB = BC. (b) Since AB = BC and opposite sides of a parallelogram are equal, AB = BC = CD = DA, so ABCD is a rhombus. The diagonals of a rhombus bisect each other at right angles, so AC and BD bisect each other at right angles.

ICSE 2027 sample paper · 3 marks

In a circle with centre O, PA and PB are tangents drawn from an external point P. If PA = 9 cm and ∠ APB = 60°, find the: (a) length of PB. (b) radius of the circle.

Show answer

(a) Tangents from an external point are equal: PB = PA = 9 cm. (b) OP bisects ∠APB, so ∠APO = 30°, and ∠OAP = 90° (radius ⊥ tangent). In △OAP: (OA)/(PA) = tan 30° = (1)/(√(3)) ⇒ OA = (9)/(√(3)) = 3√(3) = 5.196 ≈ 5.2 cm.

ICSE 2026 sample paper · 3 marks

The parallelogram ABCD circumscribe a circle, touching circle at P, Q, R and S (P on AB, Q on BC, R on CD and S on DA). (a) Prove that: AB = BC (b) What special name can be given to the parallelogram ABCD?

Show answer

(a) Tangents drawn to a circle from an external point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS. Adding: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), i.e. AB + CD = AD + BC. In a parallelogram CD = AB and AD = BC, so 2AB = 2BC ⟹ AB = BC. (b) A parallelogram with adjacent sides equal is a rhombus.

ICSE 2025 sample paper · 1 mark

O is the centre of the circle and PT is a tangent at P. QP is a diameter of the circle (Q, O and P lie on a straight line), OT is joined and ∠QOT = 110°. If ∠PTO = x°, the value of x is:

20°
40°
55°
70°

Show answer

20°

ICSE 2025 sample paper · 3 marks

PT is a tangent to the circle at T. Chord BA produced meets the tangent PT at P. TA and TB are joined. Given PT=20cm and PA= 16cm. (a) Prove ΔPTB ~ΔPAT (b) Find the length of AB.

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(a) In ΔPTB and ΔPAT: ∠PTA = ∠PBT (angle between tangent and chord equals the angle in the alternate segment) ∠TPA = ∠BPT (common angle) ∴ ΔPTB ~ ΔPAT (AA similarity). (b) From the similarity, (PT)/(PA) = (PB)/(PT), so PA × PB = PT². 16 × (16 + AB) = 20² = 400 ⇒ 16 + AB = 25 ⇒ AB = 9 cm.

ICSE 2024 sample paper · 1 mark

In a circle with radius R, the shortest distance between two parallel tangents is equal to:

R
2R
2πR
πR

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ICSE 2024 board paper · 1 mark

PS and PT are the tangents to the circle. SQ || PT and ∠ SPT = 80°. The value of ∠ QST is:

140°
90°
80°
50°

Show answer

50°