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NCERT Solutions for Class 10 Maths Chapter 10 Circles

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Edited By Ramraj Saini | Updated on Sep 07, 2023 05:18 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 10 Circles

NCERT Solutions for Class 10 Maths Chapter 10 Circles provided here. The ncert solutions have been carefully created by Maths experts at Careers360 keeping in mind of latest syllabus and pattern of CBSE 2023-24. Circles class 10 solutions are a vital resource for students preparing for the Class 10 board exams. These solutions provide step-by-step, comprehensive and detailed answers for all questions in the NCERT Class 10 Maths textbook and are arranged in an exercise-wise format to aid students' preparation and understanding. It is advisable to complete the NCERT Class 10 Maths syllabus at the earliest. To achieve high marks on the exam, students are encouraged to use these NCERT Solutions chapter 10 maths class 10 as part of their study materials. They are free to download using the link given below in this page.

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NCERT Solutions for Class 10 Maths Chapter 10 Circles Free PDF Download

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NCERT Solutions for Class 10 Maths Chapter 10 Circles - Important Points

When a circle with equation x2 + y2 = a2 is intersected by a line y = mx + c, the equation of Tangent Line to Circle: y = mx ± a √(1 + m2)

Relationship between Tangent and Radius - The tangent to a circle is perpendicular to the radius through the point of contact.

Length of Tangent from External Point: 2√(r2 - d2)

Where r is the circle's radius and d is the distance from the external point to the circle's center.

Free download NCERT Solutions for Class 10 Maths Chapter 10 Circles PDF for CBSE Exam.

NCERT Solutions for Class 10 Maths Chapter 10 Circles (Intext Questions and Exercise)

Circles class 10 NCERT solutions 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.

15942926237221594292621263

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

15942929967351594292993758
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:

15942930944311594293092133
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 circles class 10 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} 1648791076455 OT = 7 cm

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

1648791138639

(A) 60 \degree

(B) 70 \degree

(C) 80 \degree

(D) 9 0 \degree

Answer:

The correct option is (b)

15942932969511594293294689
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)
15942933816021594293378856
It is given that, tangent PA and PB from point P inclined at \angle APB = 80^0
In triangle \Delta OAP and \Delta OBP
\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, \angle OPA = 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:

15942935437651594293541307
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 \because 1 & \angle 2 are alternate angles}

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

Answer:

15942936565601594293654097
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,
\angle BXO + \angle YXB = 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:

15942937663561594293764198
Given that,
the length of the tangent from the point A (AP) is 4 cm and the length of OA is 5 cm.
Since \angle APO = 90 0
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:

15942939015431594293899175
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 \Delta OPR and \Delta OQR
\angle PRO = \angle QRO {both 90^0 }

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

By RHS congruence \Delta OPR \cong \Delta OQR
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

15942940162051594294012805

Answer:

15942941521951594294149641
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)
\Rightarrow AB + 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:

15942942510281594294247906
To prove- \angle AOB = 90^0
Proof-
In \Delta AOP and \Delta AOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence, \Delta AOP \cong \Delta AOC
and by CPCT, \angle PAO = \angle OAC
\Rightarrow \angle PAC = 2\angle OAC ..................(i)

Similarly, from \Delta OBC and \Delta OBQ, we get;
\angle QBC = 2. \angle OBC.............(ii)

Adding eq (1) and eq (2)

\angle PAC + \angle QBC = 180
2( \angle OBC + \angle OAC) = 180
( \angle OBC + \angle OAC) = 90

Now, in \Delta OAB,
Sum of interior angle is 180.
So, \angle OBC + \angle OAC + \angle AOB = 180
\therefore \angle AOB = 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:

15942943804671594294377915
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,
\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^0
90 + \angle APB + 90 + \angle BOA = 360
\angle APB + \angle AOB = 180^0
Hence proved
.

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

Answer:

15942946633761594294660855
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)]
\Rightarrow AB = 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:

1639543596459

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:

15942948311321594294828094
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 \Delta DOS and \Delta DOR,
OD =OD [common]
OS = OR [radii of same circle]
DR = DS [length of tangents drawn from an external point are equal ]
By SSS congruency, \Delta DOS \cong \Delta DOR,
and by CPCT, \angle DOS = \angle DOR
\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.

Circles Class 10 NCERT Topics

  • Tangent to a Circle
  • Radius and Tangent Relation
  • Tangent From External Point to the Circle
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Circles Class 10 Solutions - Exercise Wise

More About Circles Class 10 NCERT Solutions

NCERT solutions for circle class 10 are solved by subject experts to help students in their preparation keeping step by step marking in the mind. Circles Class 10 introduces some complex and important terms like tangents, tangents to a circle, number of tangents from a point on a circle. Class 10 Maths ch 10, we will study the different conditions that arise when a line and a circle are given in a plane. Circle class 10 has fundamental concepts that are important for students in their future studies.

NCERT solutions for class 10 Maths - Chapter Wise

NCERT Exemplar solutions - Subject wise

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How To Use NCERT Solutions For Class 10 Maths Chapter 10 Circles?

  • Circle is the chapter, the logic and the 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 solve the practice exercises of Circles Class 10 where your understanding of concepts will be tested.

  • While solving the practice exercises, you can take the help of NCERT solutions for Class 10 Maths hapter 10 circles.

  • Lastly, solve the previous years question papers related to the particular chapter.

NCERT Solutions for Class 10 - Subject Wise

Key Features of NCERT Solutions for Class 10 Maths Chapter 10 Circles

The NCERT Solutions for class 10 maths chapter 10 Circles include several key features that help students to improve their understanding and practice of the material.

  • These circle class 10 solutions include answers for all exercise questions in the NCERT Class 10 Maths textbook.

  • These solutions help students to practice challenging questions, align with the updated CBSE syllabus and cover possible exam questions, provide guidance for important drawings.

  • These solutions aid students in memorizing important formulas and calculation methods.

NCERT Books and NCERT Syllabus

Happy study!

Frequently Asked Question (FAQs)

1. What is the weightage of chapter Circles in CBSE 10 board final exam ?

Triangles, circles and constructions combined have 15 marks weightage in Class 10 final board exam. To obtain a good score in the CBSE 10th board exam follow NCERT book and the previous board papers. More questions can be practiced from NCERT exemplar. 

2. What are the main concepts covered in NCERT Solutions for class 10 maths circles?

The key concepts covered in NCERT Solutions for circle chapter class 10 include the introduction to circles, the properties of tangents to a circle, the calculation of the number of tangents from a point on a circle, and a summary of the entire chapter. Students can also use chapter 10 class 10 maths solutions in PDF format.

3. How is the NCERT solution helpful in the CBSE board exams?

As most of the questions in CBSE board exam are directly asked from NCERT textbook, so must solve all the problems given in the NCERT textbook. NCERT solutions are not only important when you are stuck while solving the problems but you will get to know how to answer in the board exam in order to get good marks in the board exam.

4. Can you list the key features of circles class 10 ncert solutions?

The NCERT Solutions for chapter 10 class 10 maths includes answers to all exercise questions in the NCERT Class 10 Maths textbook. Additionally, they provide guidance for practicing challenging questions, which will enhance students' comprehension of the topics covered in the chapter.

5. Is it mandatory to study all the questions in NCERT class 10 maths chapter 10 solutions?

It is essential to study all the questions provided in NCERT Solutions for Class 10 Maths Chapter 10 as they may be present in CBSE board exams and class tests. By familiarizing yourself with these questions, you will be well-prepared for your upcoming exams. In NCERT Solutions, class 10 Maths chapter 10 question answer are comprehensively discussed. Therefore students can use these recourses to score well in the exam.

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Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

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Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

https://school.careers360.com/boards/cbse/cbse-date-sheet

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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