In This Post we are providing Chapter 8 Circle NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Circle Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Circle NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
Exercise: 10.1
1. How many tangents can a circle have?
Answer
A circle can have infinite tangents.
2. Fill in the blanks :
(i) A tangent to a circle intersects it in …………… point(s).
(ii) A line intersecting a circle in two points is called a ………….
(iii) A circle can have …………… parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called …………
Answer
(i) one
(ii) secant
(iii) two
(iv) point of contact
3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
a point Q so that OQ = 12 cm. Length PQ is :
a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) √119 cm
(B) 13 cm
(C) 8.5 cm
(D) √119 cm
Answer
The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ
By Pythagoras theorem in ΔOPQ,
OQ2 = OP2 + PQ2
⇒ (12)2 = 52 + PQ2
⇒ (12)2 = 52 + PQ2
⇒PQ2 = 144 – 25
⇒PQ2 = 119
⇒PQ = √119 cm
(D) is the correct option.
4. Draw a circle and two lines parallel to a given line such that one is a tangent and the
other, a secant to the circle.
Answer
AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.
Exercise: 10.2
In Q.1 to 3, choose the correct option and give justification.
1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Answer
The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)
By Pythagoras theorem in ΔOPQ,
OQ2 = OP2 + PQ2
⇒ (25)2 = OP2 + (24)2
⇒ (25)2 = OP2 + (24)2
⇒ OP2 = 625 – 576
⇒ OP2 = 49
⇒ OP = 7 cm
The radius of the circle is option (A) 7 cm.
2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°
2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°
Answer
OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT = 360°
⇒ ∠PTQ + 90° + 110° + 90° = 360°
OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT = 360°
⇒ ∠PTQ + 90° + 110° + 90° = 360°
⇒ ∠PTQ = 70°
∠PTQ is equal to option (B) 70°.
3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Answer
OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°
In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB = 360°
⇒ ∠AOB + 90° + 90° + 80° = 360°
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB = 360°
⇒ ∠AOB + 90° + 90° + 80° = 360°
⇒ ∠AOB = 100°
Now,
In ΔOPB and ΔOPA,AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)
Thus ∠POB = ∠POA
∠AOB = ∠POB + ∠POA
⇒ 2 ∠POA = ∠AOB
⇒ ∠POA = 100°/2 = 50°
∠POA is equal to option (A) 50°
Page No: 214
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer
∴ OB ⊥ RS and,
∴ OA ⊥ PQ
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º
From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer
Let AB be the tangent to the circle at point P with centre O.
We have to prove that PQ passes through the point O.
Suppose that PQ doesn’t passes through point O. Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
PQ intersect CD at R and also intersect AB at P.
AS, CD // AB PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)but also,
∠RPA = 90° (PQ ⊥ AB)
⇒ ∠ORP = 90°
∠ROP + ∠OPA = 180° (Co-interior angles)
⇒∠ROP + 90° = 180°
⇒∠ROP = 90°
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.
6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Answer
OA = 5cm and AB = 4 cm (Given)
In ΔABO,
By Pythagoras theorem in ΔABO,
OA2 = AB2 + BO2
⇒ 52 = 42 + BO2
⇒ BO2 = 25 – 16
⇒ BO2 = 9
⇒ BO = 3
∴ The radius of the circle is 3 cm.
7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.
Answer
AB be the chord of the larger circle which touches the smaller circle at point P.
∴ AB is tangent to the smaller circle to the point P.
⇒ OP ⊥ AB
By Pythagoras theorem in ΔOPA,
OA2 = AP2 + OP2
⇒ 52 = AP2 + 32
⇒ AP2 = 25 – 9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
∴ The length of the chord of the larger circle is 8 cm.
8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC
Answer
From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)
By Pythagoras theorem in ΔOPA,
OA2 = AP2 + OP2
⇒ 52 = AP2 + 32
⇒ AP2 = 25 – 9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
∴ The length of the chord of the larger circle is 8 cm.
8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC
Answer
From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)
BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC
9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.
Answer
We joined O and C
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)
Similarly,
ΔOQB ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°
10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Answer
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º
∴ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
ABCD is a parallelogram,
∴ AB = CD … (i)
∴ BC = AD … (ii)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
In ΔABC,
11. Prove that the parallelogram circumscribing a circle is a rhombus.
Answer
∴ AB = CD … (i)
∴ BC = AD … (ii)
From the figure, we observe that,
DR = DS (Tangents to the circle at D)CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC … (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
⇒ AB = BC … (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
⇒ AB = BC … (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.
12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
Answer
Length of two tangents drawn from the same point to the circle are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x
We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
Now semi perimeter of triangle (s) is,
⇒ 2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
⇒s = 14 + x
⇒ 2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
⇒s = 14 + x
Area of ΔABC = √s (s – a)(s – b)(s – c)
= √(14 + x) (x)(8)(6)
= √(14 + x) 48 x … (i)also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)
= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]
= 2×1/2 (4x + 24 + 32) = 56 + 4x … (ii)
Equating equation (i) and (ii) we get,
√(14 + x) 48 x = 56 + 4x
Squaring both sides,48x (14 + x) = (56 + 4x)2
⇒ 48x = [4(14 + x)]2/(14 + x)
⇒ 48x = 16 (14 + x)
⇒ 48x = 224 + 16x
⇒ 32x = 224
⇒ x = 7 cm
Hence, AB = x + 8 = 7 + 8 = 15 cmCA = 6 + x = 6 + 7 = 13 cm
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒∠1 = ∠8
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Important Links
Circle – Quick Revision Notes
Circle- Most Important Questions
Circle – Important MCQs
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