Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
Proof idea: take any point Q on the tangent other than P. Q lies outside the circle, so OQ > OP. Thus OP is the shortest distance from O to the line, and the shortest segment is the perpendicular.
Proof: in ΔOQP and ΔORP, OQ = OR (radii), OP is common, and ∠OQP = ∠ORP = 90°. So the triangles are congruent (RHS), and PQ = PR.
| Result | Statement |
|---|---|
| Theorem 10.1 | tangent ⊥ radius at the point of contact |
| Theorem 10.2 | tangents from an external point are equal |
| Tangent length | PT² = OP² − r² |
| Angle between tangents | ∠QPR = 180° − ∠QOR |
| Chord of the larger of two concentric circles touching the smaller | length = 2√(R² − r²), bisected at the point of contact |
Try each problem first, then open the solution to check your working.
A point P is 13 cm from the centre of a circle of radius 5 cm. Find the length of the tangent from P.
Given: OP = 13, r = 5
Use: PT² = OP² − r²
PT² = 169 − 25 = 144
Answer: 12 cm
The tangent from a point 17 cm from the centre has length 15 cm. Find the radius.
Use: r² = OP² − PT²
r² = 289 − 225 = 64
Answer: 8 cm
The radius of a circle is 6 cm and the tangent from P is 8 cm. How far is P from the centre?
Use: OP² = r² + PT²
OP² = 36 + 64 = 100
Answer: 10 cm
Two concentric circles have radii 13 cm and 5 cm. Find the length of the chord of the larger circle that touches the smaller.
Use: Half-chord² = R² − r²
half-chord = √(169 − 25) = 12
chord = 24
Answer: 24 cm
A chord of length 16 cm of the larger of two concentric circles touches the smaller circle of radius 6 cm. Find the larger radius.
Use: R² = r² + (half-chord)²
R² = 36 + 64 = 100
Answer: 10 cm
From a point P, tangents PA and PB are drawn to a circle. If PA = 7 cm and ∠APB = 60°, find AB.
Given: PA = PB = 7
Use: ΔPAB is isosceles with a 60° angle
Base angles are (180° − 60°)/2 = 60°, so ΔPAB is equilateral.
Answer: AB = 7 cm
Tangents PA and PB are drawn to a circle with centre O. If ∠AOB = 130°, find ∠APB.
Use: ∠APB + ∠AOB = 180°
∠APB = 50°
Answer: 50°
If the angle between two tangents from an external point is 70°, find the angle between the radii at the points of contact.
180° − 70°
Answer: 110°
Tangents PA, PB from P to a circle with centre O make ∠APB = 80°. Find ∠OPA and ∠AOP.
Use: OP bisects ∠APB; ∠OAP = 90°
∠OPA = 40°
∠AOP = 180° − 90° − 40° = 50°
Answer: ∠OPA = 40°, ∠AOP = 50°
PT is a tangent at T to a circle with centre O, and TQ is a chord with ∠OTQ = 25°. Find ∠PTQ (Q on the same side as P).
Use: ∠OTP = 90°
∠PTQ = 90° − 25°
Answer: 65°
From P, 10 cm from the centre O of a circle of radius 5 cm, tangents PA and PB are drawn. Find ∠APB.
Use: sin ∠OPA = OA/OP
sin ∠OPA = 5/10 = 1/2 ⇒ ∠OPA = 30°
∠APB = 60°
Answer: 60°
A quadrilateral ABCD circumscribes a circle. AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
Use: AB + CD = AD + BC
6 + 4 = AD + 7
Answer: AD = 3 cm
Tangents PQ and PR of length 10 cm are drawn from P. A third tangent at E on the arc nearer P meets PQ at A and PR at B. Find the perimeter of ΔPAB.
Given: PQ = PR = 10
Use: AE = AQ, BE = BR
Perimeter = PA + AE + EB + BP = PA + AQ + BR + BP = PQ + PR
Answer: 20 cm
A circle is inscribed in ΔABC touching AB, BC, CA at P, Q, R. If AB = 10 cm, BC = 8 cm and CA = 12 cm, find AP, BQ and CR.
Given: AP = AR = x, BP = BQ = y, CQ = CR = z
x + y = 10, y + z = 8, z + x = 12
Adding: x + y + z = 15
x = 15 − 8 = 7, y = 15 − 12 = 3, z = 15 − 10 = 5
Answer: AP = 7 cm, BQ = 3 cm, CR = 5 cm
A circle is inscribed in a right triangle with legs 6 cm and 8 cm. Find its radius.
Given: hypotenuse 10
Use: In a right triangle r = (a + b − c)/2
The tangent lengths from the right-angle vertex are both r (the radii there form a square).
r = (6 + 8 − 10)/2
Answer: 2 cm
A parallelogram ABCD circumscribes a circle and AB = 5 cm. Find the other sides.
Use: Circumscribed parallelogram is a rhombus
AB + CD = AD + BC and opposite sides are equal, so 2AB = 2AD.
Answer: All sides 5 cm
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Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
At every point of a circle there is exactly one tangent, and a circle has infinitely many points.
Answer: Infinitely many
Answer: (i) one (ii) secant (iii) two (iv) point of contact
OP ⊥ PQ (Theorem 10.1), so in right ΔOPQ, PQ² = OQ² − OP² = 144 − 25 = 119.
Answer: (D) √119 cm
Draw a circle with centre O and a line l. Draw the diameter perpendicular to l. At one end of this diameter, draw a line parallel to l: it is a tangent (it is perpendicular to the radius there).
Draw another line parallel to l between the centre and that tangent: it cuts the circle at two points, so it is a secant.
Answer: Constructed as above
The radius is perpendicular to the tangent at the point of contact, so r² = 25² − 24² = 625 − 576 = 49.
Answer: (A) 7 cm
∠OPT = ∠OQT = 90°. In quadrilateral OPTQ, ∠PTQ = 360° − 90° − 90° − 110° = 70°.
Answer: (B) 70°
OP bisects ∠APB, so ∠OPA = 40°. ∠OAP = 90°. In ΔOAP, ∠POA = 180° − 90° − 40° = 50°.
Answer: (A) 50°
Let AB be a diameter and l, m the tangents at A and B.
By Theorem 10.1, l ⊥ AB and m ⊥ AB.
∠ at A and ∠ at B are both 90°, which are alternate (or co-interior summing to 180°) angles made by the transversal AB. So l ∥ m.
Answer: Proved
Let the tangent touch the circle at P, and let O be the centre. By Theorem 10.1, OP ⊥ the tangent.
At P only one line can be perpendicular to the tangent. Since OP is such a line, the perpendicular at P is OP itself, which passes through O.
Answer: Proved
r² = 5² − 4² = 9 (right angle at the point of contact).
Answer: 3 cm
Let AB touch the smaller circle at P. OP ⊥ AB, so P is the mid-point of AB (perpendicular from the centre bisects a chord).
AP² = OA² − OP² = 25 − 9 = 16 ⇒ AP = 4. AB = 8.
Answer: 8 cm
Let the circle touch AB, BC, CD, DA at P, Q, R, S. Tangents from a 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.
Answer: Proved
Join OC. In ΔOPA and ΔOCA: OP = OC (radii), AP = AC (tangents from A), OA common. So the triangles are congruent and ∠PAO = ∠CAO. Hence ∠OAB = ½ ∠PAB.
Similarly ∠OBA = ½ ∠QBA.
XY ∥ X′Y′, so ∠PAB + ∠QBA = 180° (co-interior angles). Hence ∠OAB + ∠OBA = 90°.
In ΔAOB, ∠AOB = 180° − 90° = 90°.
Answer: Proved
Let PA and PB be tangents from P to a circle with centre O. ∠OAP = ∠OBP = 90° (Theorem 10.1).
In quadrilateral OAPB, the angles add to 360°: ∠APB + ∠AOB + 90° + 90° = 360°.
So ∠APB + ∠AOB = 180°.
Answer: Proved
Let ABCD be a parallelogram circumscribing a circle. As in Q8, AB + CD = AD + BC.
In a parallelogram AB = CD and AD = BC, so 2AB = 2AD, i.e. AB = AD.
So all four sides are equal and ABCD is a rhombus.
Answer: Proved
Let the circle touch AB at F and AC at E. Tangents: BF = BD = 8, CE = CD = 6, AF = AE = x.
Sides: BC = 14, AB = x + 8, AC = x + 6. Semi-perimeter s = x + 14.
Area by Heron's formula = √[(x + 14)(x)(6)(8)] = √[48x(x + 14)].
Area also = sum of the areas of ΔOBC, ΔOCA, ΔOAB = ½ × 4 × (perimeter) = 4s = 4(x + 14).
48x(x + 14) = 16(x + 14)² ⇒ 3x = x + 14 ⇒ x = 7.
AB = 15 cm, AC = 13 cm.
Answer: AB = 15 cm, AC = 13 cm
Let ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S. Join O to the vertices and to P, Q, R, S.
ΔOAP ≅ ΔOAS (RHS), so ∠AOP = ∠AOS. Similarly ∠BOP = ∠BOQ, ∠COQ = ∠COR, ∠DOR = ∠DOS.
The eight angles at O add to 360°: 2(∠AOP + ∠BOP + ∠COR + ∠DOR) = 360°.
So (∠AOP + ∠BOP) + (∠COR + ∠DOR) = 180°, i.e. ∠AOB + ∠COD = 180°. Similarly ∠BOC + ∠AOD = 180°.
Answer: Proved