Theorems on Tangents Class 10 Concise Exe-18A ICSE Maths Selina Solutions Ch-18 Tangents and Intersecting Chords. In this article you would learn how to solve problems / questions on tangent properties.  Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Theorems on Tangents Class 10 Concise Exe-18A ICSE Maths Selina Solutions Ch-18 Tangents and Intersecting Chords

Board	ICSE
Publications	Selina
Subject	Maths
Class	10th
Chapter-18	Tangents and Intersecting Chords
Writer	R.K. Bansal
Exe-18A	Theorems on Tangents
Edition	2025-2026

Practice Questions / Problems on “Theorems on Tangents” with Solutions

Class 10 Concise Exe-18A ICSE Maths Selina Solutions Ch-18 Tangents and Intersecting Chords

Que-1: The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre?

Sol: 

OP = 10 cm; radius OT = 8 cm
∵ OT ⊥ PT
In right ΔOTP,
OP² = OT² + PT²
10² = 8² + PT²
PT² = 100 – 64
PT² = 36
PT = 6
Length of tangent = 6 cm.

Que-2: In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

Sol: Let OD = OC = x cm (radius of same circle)
Since ACD is a secant and AB is a tangent to the given circle, we have,
AC . AD = AB²
(7.5)(7.5 + 2x) = 15²
⇒ 56.25 + 15x = 225
⇒ 15x = 168.75 ⇒ x = 11.25
Thus, the radius of the circle is 11.25 cm

Que-3: Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

Sol: From Q, QA and QP are two tangents to the circle with centre O
Therefore, QA = QP  …(i)
Similarly, from Q, QB and QP are two tangents to the circle with centre O’
Therefore, QB = QP …(ii)
From (i) and (ii)
QA = QB
Therefore, tangents QA and QB are equal.

Que-4: Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.

Sol: 

From Q, QA, and QP are two tangents to the circle with centre O
Therefore, QA = QP    …(i)
Similarly, from Q, QB, and QP are two tangents to the circle with centre O.
Therefore, QB = QP …(ii)
From (i) and (ii)
QA = QB
Therefore, tangents QA and QB are equal.

Que-5: Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.

Sol: OS = 5 cm
OT = 3 cm
In Rt. Triangle OST
By Pythagoras Theorem,
ST² = OS² – OT²
ST² = 25 – 9
ST² = 16
ST = 4 cm
Since OT is perpendicular to SP and OT bisects chord SP
So, SP = 8 cm

Que-6: Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.

Sol: 

AB = 6 cm, AC = 8 cm and BC = 9 cm
Let radii of the circles having centers A, B and C be r₁, r₂ and r₃ respectively.
r₁ + r₃ = 8
r₃ + r₂ = 9
r₂ + r₁ = 6
Adding
r₁ + r₃ + r₃ + r₂ + r₂ + r₁ = 8 + 9 + 6
2(r₁ + r₂ + r₃) = 23
r₁ + r₂ + r₃ = 11.5 cm
r₁ + 9 = 11.5  …(Since r₂ + r₃ = 9)
r₁ = 2.5 cm
r₂ + 6 = 11.5  …(Since r₁ + r₃ = 6)
r₂ = 5.5 cm
r₃ + 8 = 11.5  …(Since r₂ + r₁ = 8)
r₃ = 3.5 cm
Hence, r₁ = 2.5 cm, r₂ = 5.5 cm and r₃ = 3.5 cm

Que-7: If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.

Sol: 

Let the sides of the quadrilateral ABCD touch the circle at points P, Q, R and S as shown in the figure.
We know that, tangents drawn from an external point to the circle are equal in length.
Therefore,
𝐴𝑃 = 𝐴⁢𝑆
𝐵⁢𝑃 = 𝐵⁢𝑄
𝐶⁢𝑄 = 𝐶⁢𝑅
𝐷⁢𝑅 = 𝐷⁢𝑆           …………….(1)
∴AB + CD = (AP + BP) + (CR + DR)
= (AS + BQ) + (CQ + DS) [Using (1)]
= (AS + DS) + (BQ + CQ)
= AD + BC
Hence, AB + CD = AD + BC

Que-8: If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.

Sol: 

From A, AP and AS are tangents to the circle.
Therefore, AP = AS…….(i)
Similarly, we can prove that:
BP = BQ ………(ii)
CR = CQ ………(iii)
DR = DS ………(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
But AB = CD and BC = AD…….(v) Opposite sides of a ||gm
Therefore, AB + AB = BC + BC
2AB = 2 BC
AB = BC ……..(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus.

Que-9: From the given figure prove that:
AP + BQ + CR = BP + CQ + AR.

Also, show that AP + BQ + CR = 1/2  × perimeter of triangle ABC.

Sol: Since from B, BQ and BP are the tangents to the circle
Therefore, BQ = BP   …(i)
Similarly, we can prove that
AP = AR  …(ii)
And CR = CQ  …(iii)
Adding,
AP + BQ + CR = BP + CQ + AR  …(iv)
Adding AP + BQ + CR to both sides
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR
2(AP + BQ + CR) = AB + BC + CA
Therefore, AP + BQ + CR = 1/2 × (AB + BC + CA)
AP + BQ + CR = 1/2 × perimeter of triangle ABC

Que-10: In the figure, if AB = AC then prove that BQ = CQ.

Sol: Since, from A, AP and AR are the tangents to the circle
Therefore, AP = AR
Similarly, we can prove that
BP = BQ and CR = CQ
Adding,
AP + BP + CQ = AR + BQ + CR
(AP + BP) + CQ = (AR + CR) + BQ
AB + CQ = AC + BQ
But AB = AC
Therefore, CQ = BQ or BQ = CQ

Que-11: Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if –
(i) they touch each other externally.
(ii) they touch each other internally.

Sol: Radius of bigger circle = 6.3 cm
And radius of smaller circle = 3.6 cm
(i) Two circles are touching each other at P externally. O and O’ are the centers of the circles. Join
OP and O’P
OP = 6.3 cm, O’P = 3.6 cm
Adding,
OP + O’P = 6.3 + 3.6 = 9.9 cm

(ii) Two circles are touching each other at P internally. O and O’ are the centers of the circles. Join
OP and O’P
OP = 6.3 cm, O’P = 3.6 cm
OO’ = OP – O’P = 6.3 – 3.6 = 2.7 cm

Que-12: From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
i) ∠AOP = ∠BOP
ii) OP is the perpendicular bisector of chord AB.

Sol: 

(i) In ΔAOP and ΔBOP
AP = BP   …(Tangents from P to the circle)
OP = OP   …(Common)
OA = OB   …(Radii of the same circle)
∴ By Side – Side – Side criterion of congruence,
ΔAOP ≅ ΔBOP
The corresponding parts of the congruent triangle are congruent
⇒ ∠AOP = ∠BOP  …[By c.p.c.t]

(ii) In ΔOAM and ΔOBM
OA = OB  …(Radii of the same circle)
∠AOM = ∠BOM  …(Proved ∠AOP = ∠BOP)
OM = OM  …(Common)
∴ By Side – Angle – Side criterion of congruence,
ΔOAM ≅ ΔOBM
The corresponding parts of the congruent triangles are congruent.
⇒ AM = MB
And ∠OMA = ∠OMB
But, ∠OMA + ∠OMB = 180°
∴ ∠OMA = ∠OMB = 90°
Hence, OM or OP is the perpendicular bisector of chord AB.

Que-13: In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that:
i) tangent at point P bisects AB.
ii) Angle APB = 90°

Sol: 

Draw TPT’ as common tangent to the circles.
(i) TA and TP are the tangents to the circle with centre O.
Therefore, TA = TP   …(i)
Similarly, TP = TB  …(ii)
From (i) and (ii)
TA = TB
Therefore, TPT’ is the bisector of AB.

(ii) Now in ΔATP,
∴ ∠TAP = ∠TPA
Similarly in ΔBTP, ∠TBP = ∠TPB
Adding,
∠TAP +∠TBP = ∠APB
But
∴ TAP + ∠TBP + ∠APB = 180°
⇒ ∠APB = ∠TAP + ∠TBP = 90°

Que-14: Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
∠PAQ = 2∠OPQ

Sol:  In quadrilateral OPAQ,
∠OPA = ∠OQA = 90°
(∵ OP ⊥ PA and OQ ⊥ QA)
∴ ∠POQ + ∠PAQ + 90° + 90° = 360°
⇒ ∠POQ + ∠PAQ = 360° – 180° = 180°   …(i)
In triangle OPQ,
OP = OQ  …(Radii of the same circle)
∴ OPQ = ∠OQP
But ∠POQ + ∠OPQ + ∠OQP = 180°
⇒ ∠POQ + ∠OPQ + ∠OPQ = 180°
⇒ ∠POQ + 2∠OPQ = 180°   …(ii)
From (i) and (ii)
∠POQ + ∠PAQ = ∠POQ + 2∠OPQ
⇒ ∠PAQ = 2∠OPQ

Que-15: Two parallel tangents of a circle meet a third tangent at point P and Q. Prove that PQ subtends a right angle at the centre.

Sol: 

Join OP, OQ, OA, OB and OC.
In ΔOAP and ΔOCP
OA = OC  …(Radii of the same circle)
OP = OP  …(Common)
PA = PC   …(Tangents from P)
∴ By side – side – side criterion of congruence,
ΔOAP ≅ ΔOCP  …(SSS postulate)
The corresponding parts of the congruent triangles are congruent.
⇒ ∠APO = ∠CPO     (c.p.c.t) …(i)
Similarly, we can prove that
∴ ΔOCQ ≅ ΔOBQ
⇒ ∠CQO = ∠BQO   …(ii)
∴ ∠APC = 2∠CPO and ∠CQB = 2∠CQO
But,
∠APC = ∠CQB  = 180°
(Sum of interior angles of a transversal)
∴ 2∠CPO + 2∠CQO = 180°
⇒ ∠CPO + ∠CQO = 90°
Now in ΔPOQ,
∠CPO + ∠CQO + ∠POQ = 180°
⇒ 90° + ∠POQ  = 180°
∴ ∠POQ = 90°

Que-16: ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.

Sol:  In ΔABC, ∠B = 90°
OL ⊥ AB, OM ⊥ BC and ON ⊥ AC
LBNO is a square
LB = BN = OL = OM = ON = x
∴ AL = 12 – x
∴ AL = AN = 12 – x
Since ABC is a right triangle
AC² = AB² + BC²
⇒ 13² = 12² + BC²
⇒ 169 = 144 + BC²
⇒ BC² = 25
⇒ BC = 5
∴ MC = 5 – x
But CM = CN
∴ CN = 5 – x
Now, AC = AN + NC
13 = (12 – x) + (5 – x)
13 = 17 – 2x
2x = 4
x = 2 cm

Que-17: In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
i) ∠QOR
ii) ∠QPR
given that  ∠A = 60°

Sol: The incircle touches the sides of the triangle ABC and OP ⊥  BC, OQ ⊥ AC, OR ⊥ AB
(i) In quadrilateral AROQ,
∠ORA = 90°, ∠OQA = 90°, ∠A = 60°
∠QOR = 360° – (90° + 90° + 60°)
∠QOR = 360° – 240°
∠QOR = 120°

(ii) Now arc RQ subtends ∠QOR at the centre and ∠QPR at the remaining part of the circle.
∴ ∠𝑄⁢𝑃⁢𝑅 = 1/2 ∠𝑄⁢𝑂⁢𝑅
⇒∠𝑄⁢𝑃⁢𝑅 = 1/2 × 120°
⇒ ∠QPR = 60°

Que-18: In the following figure, PQ and PR are tangents to the circle, with centre O. If , calculate:
i) ∠QOR
ii) ∠OQR
iii) ∠QSR

Sol: 

Join QR.
(i) In quadrilateral ORPQ,
OQ ⊥ OP, OR ⊥ RP
∴ ∠OQP = 90°, ∠ORP = 90°, ∠QPR = 60°
∠QOR = 360° – (90° + 90° + 60°)
∠QOR = 360° – 240°
∠QOR = 120°

(ii) In ΔQOR,
OQ = QR  …(Radii of the same circle)
∴ ∠OQR = ∠QRO  …(i)
But, ∠OQR + ∠QRO + ∠QOR = 180°
∠OQR + ∠ QRO + 120° = 180°
∠OQR + ∠QRO = 60°
From (i)
2∠OQR = 60°
∠OQR = 30°

(iii) Now arc RQ subtends ∠QOR at the centre and ∠QSR at the remaining part of the circle.
∴ ∠𝑄⁢𝑆⁢𝑅 = 1/2∠𝑄⁢𝑂⁢𝑅
⇒∠𝑄⁢𝑆⁢𝑅 = 1/2 × 120°
⇒ ∠QSR = 60°

Que-19: In the given figure, AB is a diameter of the circle, with centre O, and AT is a tangent. Calculate the numerical value of x.

Sol: In ΔOBC,
OB = OC  …(Radii of the same circle)
∴ ∠OBC = ∠OCB
But, Ext. ∠COA = ∠OBC + ∠OCB
Ext. ∠COA = 2∠OBC
⇒ 64° = 2∠OBC
∴ ∠OBC = 64∘2 = 32°
Now in ΔABT,
∠BAT = 90°  …(OA ⊥ AT)
∠OBC or ∠ABT = 32°
∴ ∠BAT + ∠ABT + x° = 180°
⇒ 90° + 32° + x° = 180°
⇒ 122° + x° = 180°
⇒ x° = 180° – 122°
⇒ x° = 58°

Que-20: In quadrilateral ABCD, angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle.

Sol: BQ and BR are the tangents from B to the circle.
Therefore, BR = BQ = 27 cm.
Also RC = (38 − 27) = 11 cm
Since CR and CS are the tangents from C to the circle
Therefore, CS = CR = 11 cm
So, DS = (25 − 11) = 14 cm
Now DS and DP are the tangents to the circle
Therefore, DS = DP
Now, ∠PDS = 90°  …(Given)
And OP ⊥ AD, OS ⊥ DC
Therefore, radius = DS = 14 cm

Que-21: In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given  and ∠SPR = x° and ∠QRP = y°
Prove that -;
i) ∠ORS = y°
ii) write an expression connecting x and y

Sol: ∠QRP = ∠OSR = y  …(Angles in the alternate segment)
But OS = OR  …(Radii of the same circle)
∴ ∠ORS = ∠OSR = y°
∴ OQ = OR  …(Radii of the same circle)
∴ ∠OQR = ∠ORQ = 90° – y°  …(i) (Since OR ⊥ PT)
But in ΔPQR,
Ext. ∠OQR = x° + y°  …(i)
From (i) and (ii)
x° + y° = 90° – y°
⇒ x° + 2y° = 90°

Que-22: PT is a tangent to the circle at T. If ; calculate:
i) ∠CBT
ii) ∠BAT
iii) ∠APT

Sol: 

Join AT and BT.
(i) TC is the diameter of the circle
∴ ∠CBT = 90°   …(Angle in a semi-circle)

(ii) ∠CBA = 70°
∴ ∠ABT = ∠CBT – ∠CBA
= 90° – 70°
= 20°
Now, ∠ACT = ∠ABT = 20°  …(Angle in the same segment of the circle)
∴ ∠TCB = ∠ACB – ∠ACT
= 50° – 20°
= 30°
But, ∠TCB = ∠TAB   …(Angles in the same segment of the circle)
∴ ∠TAB or ∠BAT = 30°

(iii) ∠BTX = ∠TCB = 30°  …(Angles in the same segment)
∴ ∠PTB = 180° – 30°
∴ ∠PTB = 150°
Now in ΔPTB
∠APT + ∠PTB + ∠ABT = 180°
⇒ ∠APT + 150° + 20° = 180°
⇒ ∠APT = 180° – (150° + 20°)
⇒ ∠APT = 180° – 170°
⇒ ∠APT = 10°

Que-23: In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.

Sol: 

Join OC.
Therefore, PA and PC are the tangents
∴ OA ⊥ PA and OC ⊥ PC
In quadrilateral APCO,
∠APC + ∠AOC = 180°
⇒ 80° + ∠AOC = 180°
⇒ ∠AOC = 100°
∠BOC = 360° – (∠AOB + ∠AOC)
∠BOC =  360° – (140° + 100°)
∠BOC = 360° – 240° = 120°
Now, arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.
∴ ∠𝐵⁢𝐴⁢𝐶 = 1/2 ∠𝐵⁢𝑂⁢𝐶
∠𝐵⁢𝐴⁢𝐶 = 1/2 × 120°
= 60°

Que-24: In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.

Sol:

∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ)
∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)
∠CAQ + ∠PAC = 180°……(angles in linear pair)
∴∠PAC = 120°
∠PAC = 2∠CAD……(AD is angle bisector of ∠PAC)
∠CAD = 60°
Now,
∠CAD + ∠CAB = 60 + 30 = 90°
∠DAB = 90°
Thus, BD subtends 90° on the circle
So, BD is the diameter of circle

–: Theorems on Tangents Class 10 Concise Exe-18A ICSE Maths Selina Solutions :–

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