NCERT Solutions For Class 10th Maths Chapter 10 : Circles

CBSE NCERT Solutions For Class 10th Maths Chapter 10 : Circles. NCERT Solutins For Class 10 Mathematics. Exercise 10.1, Exercise 10.2.


NCERT Solutions for Class X Maths Chapter 10 Circles – Mathematics CBSE

Page No: 209

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) 12 cm            (B) 13 cm          (C) 8.5 cm         (D) √119 cm

Answer

https://3.bp.blogspot.com/-w07IWkxFsus/VX2WnmKmK1I/AAAAAAAAAys/ACqfEBn2j_c/s1600/ch10-circles-class10-1.png

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)= 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

https://1.bp.blogspot.com/-VhdzcPTUmkw/VX2hHMFPvZI/AAAAAAAAAy8/J95hpbADmWc/s1600/ch10-circles-class10-2.png

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.

Page NO: 213

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

Answer

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

https://3.bp.blogspot.com/-0YKKvimFPRM/VX5miOGUzfI/AAAAAAAAAzQ/2rUevIYUZmg/s1600/ch10-circles-class10-maths-1.png

∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)

By Pythagoras theorem in ΔOPQ,

OQ2 = OP2 + PQ2
⇒ (25)= OP2 + (24)2

⇒OP2 = 625 – 576

⇒OP2 = 49

⇒OP = 7 cm

The radius of the circle is option (A) 7 cm.

https://1.bp.blogspot.com/-BBi9zWc_WZA/VX5n_mByxiI/AAAAAAAAAzg/2vOMaBp63pE/s1600/ch10-circles-class10-maths-10.11.png
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°

⇒ ∠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°

https://4.bp.blogspot.com/-RfSPcljCEuQ/VX5qRRYGlrI/AAAAAAAAAzo/Q_6CwtsrCKo/s1600/ch10-circles-class10-maths-3.pngAnswer

OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ 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°

⇒ ∠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

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B

https://4.bp.blogspot.com/-dPHe4eVgu_E/VX5vgaOG5dI/AAAAAAAAAz4/N67s9Qk1O54/s1600/ch10-circles-class10-maths-4.png

respectively.
Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and,

∴ OA ⊥ PQ
∠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.

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

https://1.bp.blogspot.com/-tFrjHC0ws-8/VX5x_52N_nI/AAAAAAAAA0E/0_gbQICJ2_o/s1600/ch10-circles-class10-maths-5.png

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

AB is a tangent drawn on this circle from point A.

https://1.bp.blogspot.com/-54eXNxw5NyU/VX52N6M2MEI/AAAAAAAAA0Q/V9LvZzOAaCk/s1600/ch10-circles-class10-maths-6.png

∴ OB ⊥ AB
OA = 5cm and AB = 4 cm (Given)
In ΔABO,
By Pythagoras theorem in ΔABO,
OA2 = AB+ BO2
⇒ 5= 4+ 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

https://3.bp.blogspot.com/-XnVLruyZ01k/VX79h1ZJAII/AAAAAAAAA0k/komFxGBv5Vw/s1600/ch10-circles-class10-maths-7.png

Let the two concentric circles with centre O.

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

https://4.bp.blogspot.com/-GLcxcbOU90g/VX8UvB72GII/AAAAAAAAA00/dmCZbErQDBM/s1600/ch10-circles-class10-maths-10.12.png
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)

⇒ 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

https://1.bp.blogspot.com/-ez34RPEvelA/VX8YkZyE7kI/AAAAAAAAA1A/xqNbYreZoLc/s1600/ch10-circles-class10-maths-10.13.png

A/q,
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°

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

https://4.bp.blogspot.com/-JTCUYpyc-2o/VX8Z_UVRZJI/AAAAAAAAA1M/pi-AOamBVAs/s1600/ch10-circles-class10-maths-10.png

Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
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.

11. Prove that the parallelogram circumscribing a circle is a rhombus.

Answer

https://3.bp.blogspot.com/-qpBNMkMc44w/VX8a6jEXVkI/AAAAAAAAA1Y/-PEiRVRr_aE/s1600/ch10-circles-class10-maths-11.png

ABCD is a parallelogram,
∴ 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.

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

https://2.bp.blogspot.com/-IdDJY2iBjk0/VX8a6nG_5II/AAAAAAAAA1s/nUOwo4NGdk4/s1600/ch10-circles-class10-maths-12-1.png

In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,
∴ 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

Now semi perimeter of circle s,
⇒ 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) (14 +  14)(14 +  x – 6)(14 +  x – 8)

= √(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 (4+ 24 + 32) = 56 + 4… (ii)

Equating equation (i) and (ii) we get,

√(14 + x) 48 = 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 cm
CA = 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

https://4.bp.blogspot.com/-XBAzvlEehs0/VX8a60In-bI/AAAAAAAAA1o/PJGo4xp_O0w/s1600/ch10-circles-class10-maths-13.png

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
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

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.


GO BACK TO CLASS X MATHS ALL CHAPTERS SOLUTION


Advertisements

One thought on “NCERT Solutions For Class 10th Maths Chapter 10 : Circles

  1. Hi there friends, how is everything, in my view its genuinely awesome in favor
    of me.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.