Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  NCERT Solutions: Circles (Exercise 10.1 & 10.2)

NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

Exercise 10.1

Q1. How many tangents can a circle have?
Sol. 
A circle can have an infinite number of tangents.
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

Q2. 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 ...........

Sol. 
(i) Exactly one
(ii) secant
(iii) two
(iv) point of contact.

Q3. 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 of PQ is:
(a) 12 cm
(b) 13 cm
(c) 8.5 cm

(d) √119 cm

Ans. (d)
Sol. 
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

Radius of the circle = 5 cm
OQ = 12 cm
∠OPQ = 900
[The tangent to a circle is perpendicular to the radius through the point of contact]
PQ2 = OQ2 - OP[By Pythagoras theorem]
PQ= 12- 5= 144 - 25 = 119
PQ = √119 cm
Hence correct option is (d)


Q4. 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.
Sol. A line 'm' is parallel to the given line 'n' and a line 'l' which is secant is parallel to the given line 'n'.
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

Exercise 10.2

In Q.1 to 3, choose the correct option and give justification.

Q1. 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
Sol. From figure
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1) 
OQ2 = OP2 + PQ2
(25)2 = OP+ (24)2
⇒ 625 - 576 = OP2
⇒ 49 = OP2
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
OP = 7cm
Radius of the circle = 7cm.
Hence, correct option is (a).

Q2. In figure, 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°

NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
Sol. ∠OPT = 90°; ∠OQT = 90°; ∠POQ = 1100
TPOQ is a quadrilateral
⇒ ∠PTQ + ∠POQ = 180°
⇒ ∠PTQ + 1100 = 1800
⇒ ∠PTQ  = 1800 - 1000 = 700

Hence, correct option is (b).

Q3. Choose the correct option: If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to
(A) 50° 
(B) 60° 
(C) 70° 
(D) 80°
Sol. In ΔOAP and ΔOBP
OA = OB    [Radii]
PA = PB
[Length of tangents from an external point are equal]
OP = OP     [common]
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
Hence, Option (a) is correct

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Sol. AB is the diameter of the circle, p and q are two tangents
OA ⊥ p and OB ⊥ q and ∠1 = ∠2 = 900
⇒ p q    [∠1 and ∠2 are alternate angles]
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Sol. XY tangent to the circle C(O, r) at B and AB ⊥ XY
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
Join OB
∠ABY = 90º   
∠OBY = 90º
[Radius through the point of contact is ⊥ to the tangent]
∴ ∠ABY + ∠OBY = 1800
⇒ ABO is collinear.
∴ AB passes through the centre of the circle.

Q6. 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.
Sol. OP = Radius of the circle
OA = 5 cm; AP = 4 cm
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

∠OPA = 90°   [Radius and tangent are ⊥ar]
OA2 = AP2 + OP2    [By Pythagoras theorem]
52 = 42 + OP2 
⇒ 25 = 16 + OP2
⇒ 25 - 16 = OP2
⇒ 9 = OP2
⇒ OP = √ 9 = 3
Radius = 3 cm

Q7. 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.
Sol. The radius of larger circle = 5 cm and the radius of a smaller circle = 3 cm
OP ⊥ AB    [Radius of the circle is perpendicular to the tangent]
AB is a chord of the larger circle
∴ OP bisect  AB   AP = BP
In ΔOAP     OA2 =    AP2 + OP2
    (5)2 =   AP2 + (3)2
    AP2 =   25 - 9 = 16
  NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
AB = 2AP =   2 x 4 = 8 cm

Q8. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that: AB + CD = AD + BC

NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

Sol. 
AP = AS    ...(i)    [Lengths of tangents from an external point are equal]
BP = BQ    ...(ii)
CR = CQ ...(iii)
DR = DS   ...(iv)
Adding equations (i), (ii), (iii) and (iv), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) +(BQ + CQ
 AB + CD = AD + BC

Q9. In the figure, 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°.

NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
Sol. Given: Two parallel tangents to a circle with centre O. Tangent AB with point of contact C intersects XY at A and X'Y' at B
To Prove: ∠AOB = 90°
Construction: Join OA, OB and OC
Proof: In ΔAOP and ΔAOC
AP = AC [Lengths of tangents]
OP = OC [Radii]
OA = OA
⇒ ΔAOP ≅ ΔAOC   [SSS congruence rule]
⇒ ∠PAO = ∠CAO    [C.P.C.T]
∠PAC = 2 ∠OAC    ...(i)
Similarly ∠QBC = 2 ∠OBC    ...(ii)
Adding (i) and (ii), we get
∠PAC + ∠QBC = 2      [∠OAC + ∠OBC]
∠PAC + ∠QBC = 180°
[interior consecutive angle on same side of transversal]
2 = 180° [∠OAC + ∠OBC]
⇒ ∠OAC + ∠OBC = 90°
In ΔAOB, ∠AOB + [∠OAC + ∠OBC] = 180°
⇒ ∠AOB + 90° = 180°
⇒ ∠AOB = 90°

Q10. 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.
Sol. PA and PB are two tangents, and A and B are the points of contact of the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
[Radius and tangent are perpendicular to each other]
In quadrilateral OAPB
(∠OAP + ∠OBP) + ∠APB + ∠AOB = 3600
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
⇒ 180° + ∠APB + ∠AOB = 360°
∠APB + ∠AOB = 360° - 180° = 180°

Q11. Prove that the parallelogram circumscribing a circle is a rhombus. 
Sol. Parallelogram ABCD circumscribing a circle with centre O.
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
OP⊥ AB and OS ⊥ AD
In ΔOPB and ΔOSD, ∠OPB = ∠OSD [Each 90°]
OB = OD
[Diagonals of ║gm bisect each other]
OP = OS    [Radii]
  ΔOPB ≅ ΔOSD [RHS congruence rule]
PB = SD     [C.P.C.T] ...(i)
AP = AS [Lengths of tangents]... (ii)
Adding (i) and (ii)
AP + PB = AS + DS
   AB AD
Similarly  AB = BC = CD = DA
∴ ║gm ABCD is a rhombus

Q12. 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 figure). Find the sides AB and AC.
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
Sol. BD = 8 cm and DC = 6 cm
BE = BD = 8 cm; CD= CF = 6 cm;
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
Let AE=AF = x cm
In ΔABC   a = 6 + 8 = 14 cm;
b = (x + 6) cm;
c = (x +8)
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
ar ΔABC = ar ΔOBC + ar ΔOCA + ar ΔOAB
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
From (i) and (ii)
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
⇒ 3x(x+14) = (x + 14)2 
⇒ 3x2 + 42x = x2 + 196 + 28x
⇒ 2x2 + 14x - 196 = 0
⇒ x2 + 7x - 98 = 0
⇒ x2 + 14x - 7x - 98 = 0
⇒ x(x + 14) - 7(x + 14) = 0
⇒ (x - 7)(x + 14) = 0 ⇒ x = 7
AB = x + 8 = 7 + 8 = 15 cm
AC = x + 6 = 7 + 6 = 13 cm

Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Sol. 
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

AB touches P and BC, and CD and DA touch the circle at Q, R and S.
Construction: Join OA, OB, OC, OD and OP, OO, OR, OS
∴ ∠1 = ∠2    [OA bisects ∠POS]
Similarly
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)
Similarly ∠AOB + ∠COD = 180°
Hence, opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of a circle.

The document NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1) is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
124 videos|457 docs|77 tests

Top Courses for Class 10

FAQs on NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

1. What is the standard equation of a circle?
Ans. The standard equation of a circle with center at the point \((h, k)\) and radius \(r\) is given by the formula \((x - h)^2 + (y - k)^2 = r^2\). This equation represents all the points \((x, y)\) that are at a distance \(r\) from the center \((h, k)\).
2. How can we find the radius of a circle if we know its equation?
Ans. To find the radius of a circle from its equation, you first need to express the equation in the standard form \((x - h)^2 + (y - k)^2 = r^2\). If the equation is given in the form \(x^2 + y^2 + Dx + Ey + F = 0\), you can rearrange it to find \(r\) as follows: \(r = \sqrt{\left(\frac{-D}{2}\right)^2 + \left(\frac{-E}{2}\right)^2 - F}\).
3. What is the relationship between the diameter and radius of a circle?
Ans. The diameter of a circle is twice the length of the radius. It can be expressed with the formula \(D = 2r\), where \(D\) is the diameter and \(r\) is the radius. Therefore, if you know the radius, you can easily find the diameter by multiplying the radius by 2.
4. How do you find the area of a circle?
Ans. The area \(A\) of a circle can be calculated using the formula \(A = \pi r^2\), where \(r\) is the radius of the circle and \(\pi\) is a constant approximately equal to 3.14. This formula gives you the total space enclosed within the boundary of the circle.
5. What are the different forms of the equation of a circle?
Ans. The equation of a circle can be expressed in different forms. The standard form is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. The general form is \(x^2 + y^2 + Dx + Ey + F = 0\), which can be derived from the standard form by expanding and rearranging the equation.
124 videos|457 docs|77 tests
Download as PDF
Explore Courses for Class 10 exam

Top Courses for Class 10

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

practice quizzes

,

shortcuts and tricks

,

Summary

,

ppt

,

Previous Year Questions with Solutions

,

Free

,

Semester Notes

,

Exam

,

past year papers

,

Important questions

,

NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

,

Viva Questions

,

NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

,

pdf

,

Extra Questions

,

Objective type Questions

,

Sample Paper

,

mock tests for examination

,

video lectures

,

NCERT Solutions for Class 10 Maths Chapter 10 - Circles (Exercise 10.1)

,

MCQs

,

study material

;