NCERT Solutions for Class 10 Maths Chapter 10 Circles

 

NCERT Solutions for Class 10 Maths Chapter 10 Circles - In our previous classes, you have learnt that a circle is a closed shape with a collection of points in a plane that are at a specific distance ( called the radius) from a fixed point (called center). Solutions of NCERT class 10 maths chapter 10 Circles is covering the in-depth explanations to questions related to a circle. You have also studied important terms related to the circle like segment, arc, sector, chord, etc. In this chapter, there are two exercises with 17 questions in them. CBSE NCERT solutions for class 10 maths chapter 10 Circles are solved by subject experts to help students in their preparation keeping step  by step marking in the mind. This chapter introduces some complex and important terms like tangents, tangents to a circle, number of tangents from a point on a circle. In this chapter, we will study the different conditions that arise when a line and a circle are given in a plane. To solve these types of situations, we will learn the approach to apply the concept of the tangent to a circle in NCERT solutions for class 10 maths chapter 10 Circles. This chapter has fundamental concepts that are important for students in their future studies. Circles is a very interesting chapter due to the involvement of geometrical calculations and diagrams. These NCERT solutions can be used to study for competitive exams like JEE and NEET to build the basics.

NCERT solutions for class 10 maths chapter 10 Circles Excercise: 10.1

Q1 How many tangents can a circle have?

Answer:

The lines that intersect the circle exactly at one single point are called tangents. In a circle, there can be infinitely many tangents.

Q2 Fill in the blanks :
(1)    A tangent to a circle intersects it in_________ point (s).
(2)    A line intersecting a circle in two points is called a _____.
(3)    A circle can have ________ parallel tangents at the most.
(4)    The common point of a tangent to a circle and the circle ______

Answer:

(a) one
A tangent of a circle intersects the circle exactly in one single point. 

(b) secant
It is a line that intersects the circle at two points. 

(c) Two,
There can be only two parallel tangents to a circle.

(d) point of contact
The common point of a tangent and a circle.

Q3 A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Length PQ is :

       (A) 12 cm

       (B) 13 cm

       (C) 8.5 cm

       (D) \sqrt { 119} cm.

Answer:

The correct option is (d) = \sqrt { 119} cm


It is given that the radius of the circle is 5 cm. OQ = 12 cm
According to question,
We know that \angle QPO=90^0
So, triangle OPQ is a right-angle triangle. By using Pythagoras theorem, 
PQ =\sqrt{OQ^2-OP^2}=\sqrt{144-25}
                                                =\sqrt{119} cm

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.

Answer:


AB is the given line and the line CD is the tangent to a circle at point M and parallels to the line AB. The line EF is a secant parallel to the AB

NCERT solutions for class 10 maths chapter 10 Circles Excercise: 10.2

Q1 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is

        (A) 7 cm

       (B) 12 cm

        (C) 15 cm

        (D) 24.5 cm

Answer:

The correct option is (A) = 7 cm

Given that,
The length of the tangent (QT) is 24 cm and the length of OQ is 25 cm.
Suppose the length of the radius OT be l cm.
We know that \Delta OTQ is a right angle triangle. So, by using Pythagoras theorem-

\\OQ^2 = TQ^2+OT^2\\\\ l = \sqrt{25^2-24^2}\\\\OT = l=\sqrt{49}

OT = 7 cm

Circles Excercise: 10.2

Q2 In Fig. 10.11, if TP and TQ are the two tangents to a circle with center O so that \angle POQ = 11 0 \degree, then \angle PTQ is equal to

                                                                                                             

         (A) 60 \degree

         (B) 70 \degree                                                              

         (C) 80 \degree

         (D)9 0 \degree                                                    

Answer:

The correct option is (b)


In figure, \angle POQ = 110^0
Since POQT is quadrilateral. Therefore the sum of the opposite angles are 180

\\\Rightarrow \angle PTQ +\angle POQ = 180^0\\\\\Rightarrow \angle PTQ = 180^0-\angle POQ
                 = 180^0-100^0
                  = 70^0

Q3 If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of 80 \degree, then \angle POA is equal to

      (A) 50°

      (B) 60°

      (C) 70°

      (D) 80°

Answer:

The correct option is (A)

It is given that,  tangent PA and PB from point P inclined at \angle APB = 80^0
In triangle \DeltaOAP and \DeltaOBP
\angle OAP = \angle OBP = 90
OA =OB (radii of the circle)
PA = PB (tangents of the circle)

Therefore, by  SAS congruence
\therefore \Delta OAP \cong \Delta OBP

By CPCT, \angle OPA = \angle OPB
Now, \angleOPA = 80/4 = 40

In \Delta PAO,
\angle P + \angle A + \angle O = 180
\angle O = 180 - 130
        = 50

Q4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer:


Let line p and line q are two tangents of a circle and AB is the diameter of the circle.
OA and OB are perpendicular to the tangents p and  q respectively.
therefore,
 \angle1 = \angle2 = 90^0

\Rightarrow p || q {  \angle\because1  &  \angle2 are alternate angles}

Q5 Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.

Answer:


In the above figure, the line AXB is the tangent to a circle with center O. Here, OX is the perpendicular to the tangent AXB (OX\perp AXB) at point of contact X. 
Therefore, we have,
\angleBXO + \angleYXB = 90^0+90^0=180^0

\therefore OXY is a collinear 
\Rightarrow OX is passing through the center of the circle.

Q6 The length of a tangent from a point A at distance 5 cm from the center of the circle is 4 cm. Find the radius of the circle.

Answer:


Given that,
the length of the tangent from the point A (AP) is 4 cm and the length of OA is 5 cm.
Since \angleAPO = 900
Therefore, \Delta APO is a right-angle triangle. By using Pythagoras theorem;

OA^2=AP^2+OP^2
5^2 = 4^2+OP^2
OP=\sqrt{25-16}=\sqrt{9}
OP = 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.

Answer:


In the above figure, Pq is the chord to the larger circle, which is also tangent to a smaller circle at the point of contact R.
We have,
radius of the larger circle OP = OQ = 5 cm
radius of the small circle (OR) = 3 cm

OR \perp PQ  [since PQ is tangent to a smaller circle]

According to question,

In \DeltaOPR and \DeltaOQR
\anglePRO = \angleQRO  {both 90^0}

OR = OR {common}
OP = OQ {both radii}

By RHS congruence \DeltaOPR \cong \DeltaOQR
So, by CPCT
PR = RQ
Now, In \Delta OPR,
by using pythagoras theorem,
PR = \sqrt{25-9} =\sqrt{16}
PR = 4 cm
Hence, PQ  = 2.PR =  8 cm 

Q8 A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

       Circles EXERCISE 10.2

Answer:


To prove- AB + CD = AD + BC
Proof-
We have,
Since the length of the tangents drawn from an external point to a circle are equal 
AP =AS .......(i)
BP = BQ.........(ii)
AS = AP...........(iii)
CR = CQ ...........(iv)

By adding all the equations, we get;

    AP + BP +RD+ CR = AS +DS +BQ +CQ
\Rightarrow(AP + BP) + (RD + CR) = (AS+DS)+(BQ + CQ)
\RightarrowAB + CD  = AD + BC

Hence proved.

Q9 In Fig. 10.13, XY and X'Y'are two parallel tangents to a circle with center O and another tangent AB with a point of contact C intersecting XY at A and X'Y' at B. Prove that \angle AOB = 90°.

Answer:


To prove- \angleAOB = 90^0
Proof-
In \DeltaAOP and \DeltaAOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence,  \DeltaAOP \cong \DeltaAOC
and by CPCT,  \angle PAO = \angle OAC
\Rightarrow \angle PAC = 2\angle OAC..................(i)

Similarly, from \DeltaOBC and \DeltaOBQ, we get;
\angleQBC = 2.\angleOBC.............(ii)

Adding eq (1) and eq (2)

\anglePAC + \angle QBC = 180 
2(\angleOBC + \angleOAC) = 180
(\angleOBC + \angleOAC) = 90

Now,  in \Delta OAB,
Sum of interior angle is 180.
So, \angleOBC + \angleOAC + \angleAOB = 180
\therefore \angleAOB = 90
hence proved.

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 center.

Answer:


To prove - \angle APB + \angle AOB = 180^0 
Proof-
We have, PA and PB are two tangents, B and A are the point of contacts of the tangent to a circle. And OA\perp PA , OB\perp PB (since tangents and radius are perpendiculars)

According to question,
In quadrilateral PAOB,
\angleOAP + \angleAPB +\anglePBO +\angleBOA = 360^0
90 + \angleAPB + 90 +\angleBOA = 360
\angle APB + \angle AOB = 180^0
Hence proved
.

Q11 Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:


To prove -  the parallelogram circumscribing a circle is a rhombus
Proof-
ABCD is a parallelogram that circumscribes a circle with center O.
P, Q, R, S are the points of contacts on sides AB, BC, CD, and DA  respectively
AB = CD .and AD = BC...........(i)
It is known that tangents drawn from an external point are equal in length.
RD = DS ...........(ii)
RC = QC...........(iii)
BP = BQ...........(iv)
AP = AS .............(v)
By adding eq (ii) to eq (v) we get;
(RD + RC) + (BP + AP) = (DS + AS) + (BQ + QC)
CD + AB = AD + BC
\Rightarrow 2AB = 2AD [from equation (i)]
\RightarrowAB = AD
Now, AB = AD and AB  = CD 
\therefore AB = AD = CD = BC

Hence 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 Fig. 10.14). Find the sides AB and AC.

       

Answer:

Exercise 10.2, Q. 12 - A triangle ABC is drawn to circumscribe a circle

Consider the above figure. Assume center O touches the sides AB and AC of the triangle at point E and F respectively.

Let the length of AE is x.

Now in \bigtriangleup ABC,

CF = CD = 6 (tangents on the circle from point C)

BE = BD = 6 (tangents on the circle from point B)

AE = AF = x(tangents on the circle from point A)

Now AB = AE + EB


\\AB = AE + EB\\\\ => AB = x + 8\\\\ BC = BD + DC\\\\ => BC = 8+6 = 14\\\\ CA = CF + FA\\\\ => CA = 6 + x\\\\

Now
\\s = (AB + BC + CA )/2\\\\ => s = (x + 8 + 14 + 6 +x)/2\\\\ => s = (2x + 28)/2\\\\ => s = x + 14

Area of triangle \bigtriangleup ABC

\\=\sqrt{s*(s-a)*(s-b)*(s-c)}\\\\=\sqrt{(14+x)*[(14+x)-14]*[(14+x)-(6+x)]*[(14+x)-(8+x)]}\\\\=4\sqrt{3(14x+x^2)}
Now the area of \bigtriangleup OBC

\\= (1/2)*OD*BC\\\\ = (1/2)*4*14\\\\ = 56/2 = 28
Area of\bigtriangleup OCA

\\= (1/2)*OF*AC \\\\= (1/2)*4*(6+x) \\\\= 2(6+x) \\\\= 12 + 2x

Area of \bigtriangleup OAB

\\= (1/2)*OE*AB \\\\= (1/2)*4*(8+x) \\\\= 2(8+x) \\\\= 16 + 2x

Now Area of the \bigtriangleup ABC = Area of \bigtriangleup OBC + Area of \bigtriangleup OCA + Area of \bigtriangleup OAB

\\=> 4\sqrt{3x(14+x)}= 28 + 12 + 2x + 16 + 2x \\\\=> 4\sqrt{3x(14+x)} = 56 + 4x \\\\=> 4\sqrt{3x(14+x)} = 4(14 + x) \\\\=> \sqrt{3x(14+x)}] = 14 + x

On squaring both the side, we get

\\3x(14 + x) = (14 + x)^2\\\\ => 3x = 14 + x \:\:\:\:\:\: (14 + x = 0 => x = -14\: is\: not\: possible) \\\\=> 3x - x = 14\\\\ => 2x = 14\\\\ => x = 14/2\\\\ => x = 7

Hence

AB = x + 8

=> AB = 7+8

=> AB = 15

AC = 6 + x

=> AC = 6 + 7

=> AC = 13

Answer-  AB = 15 and AC = 13

Q13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Answer:


Given- ABCD is a quadrilateral circumscribing a circle. P, Q, R, S are the point of contact on sides AB, BC, CD, and DA respectively.

To prove- 
 \\\angle AOB + \angle COD =180^0\\ \angle AOD + \angle BOC =180^0
Proof - 
Join OP, OQ, OR and OS
In triangle \DeltaDOS and \DeltaDOR,
OD =OD [common]
OS = OR [radii of same circle]
DR = DS [length of tangents drawn from an external point are equal ]
By SSS congruency, \DeltaDOS \cong \DeltaDOR,
and by CPCT, \angleDOS = \angleDOR
                          \angle c = \angle d.............(i)

Similarily,     
                    \\\angle a = \angle b\\ \angle e = \angle f\\ \angle g =\angle h...............(2, 3, 4)

 \therefore 2(\angle a +\angle e +\angle h+\angle d) = 360^0
\\(\angle a +\angle e) +(\angle h+\angle d) = 180^0\\ \angle AOB + \angle DOC = 180^0
SImilarily, \angle AOD + \angle BOC = 180^0

Hence proved.

NCERT solutions for class 10 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers

Chapter 2

NCERT solutions for class 10 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables

Chapter 4

CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations

Chapter 5

NCERT solutions for class 10 chapter 5 Arithmetic Progressions

Chapter 6

Solutions of NCERT class 10 maths chapter 6 Triangles

Chapter 7

CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry

Chapter 8

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

Chapter 9

Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry

Chapter 10

CBSE NCERT solutions class 10 maths chapter 10 Circles

Chapter 11

NCERT solutions  for class 10 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles

Chapter 13

CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes

Chapter 14

NCERT solutions for class 10 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 10 maths chapter 15 Probability

Solutions of NCERT class 10 subject wise

NCERT solutions for class 10 maths

Solutions of NCERT for class 10 science

How to use NCERT solutions for class 10 maths chapter 10 Circles?

  • It is the chapter the logic as well as memory, both are tested equally.

  • First, you should learn the theorems, terminologies, and concepts of circles.

  • Once you have done the theorems, go through some examples of the NCERT textbook.

  • When you have done the above-said points, then come to the practice exercises where your understanding of concepts will be tested.

  • While doing the practice exercises, you can take help of NCERT solutions for class 10 maths chapter 10 circles.

  • After doing all this, the last thing you should solve is past papers related to the particular chapter.

Keep working hard & happy learning!

 

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