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NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 - Conic Section

NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 - Conic Section

Edited By Vishal kumar | Updated on May 06, 2025 03:15 PM IST

Understanding the topic of circles and their equation is important, as it has a significant wide range of real-life applications, such as the use of geometry while designing wheels, designing of circular tracks and many more. Furthermore, the study of orbits and revolutions of satellites is driven by the study of circles. The first part of this chapter, 'Conic Sections', deals with various aspects of circles. These circles are formed when a plane intersects a cone in a particular way. Thereafter advanced concepts of conic sections are discussed.

In this NCERT exercise, 10.1 of Chapter 10, we have dealt with the questions related to circles, along with the representation of the equation of a circle in different forms. The exercise also involves solving questions such as determining the equation of a circle when its centre and radius are provided or identifying the centre and radius from a given equation of a circle. To get conceptual clarity on this topic, students are advised to solve the given problems on their own in the first go without seeing the solutions, thereafter referring to the solutions provided. Students can also access complete NCERT Solutions from Class 6 to Class 12 provided here in an elaborate manner.

This Story also Contains
  1. Download PDF of NCERT Solutions for Class 11 Maths Chapter 10 – Conic Sections Exercise 10.1
  2. NCERT Solutions Class 11 Maths Chapter 10: Exercise 10.1
  3. Question:1 Find the equation of the circle with
  4. Topics covered in Chapter 10 Conic Section : Exercise 10.1
  5. Class 11 Subject-Wise Solutions
  6. NCERT Solutions of Class 11 Subject Wise
  7. Subject Wise NCERT Exampler Solutions
NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 - Conic Section
NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.1 - Conic Section


Download PDF of NCERT Solutions for Class 11 Maths Chapter 10 – Conic Sections Exercise 10.1

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NCERT Solutions Class 11 Maths Chapter 10: Exercise 10.1

Question:1 Find the equation of the circle with

centre (0,2) and radius 2

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

So Given Here

(h,k)=(0,2)

AND r=2

So the equation of the circle is:

, (x0)2+(y2)2=22

x2+y24y+4=4

x2+y24y=0

Question:2 Find the equation of the circle with

centre (–2,3) and radius 4

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

So Given Here

(h,k)=(2,3)

AND r=4

So the equation of the circle is:

, (x(2))2+(y3)2=42

x2+4x+4+y26y+9=16

x2+y2+4x6y3=0

Question:3 Find the equation of the circle with

centre (12,14) and radius 112

Answer:

As we know,

The equation of the circle with center ( h, k) and radius r is give by ;

(xh)2+(yk)2=r2

So Given Here

(h,k)=(12,14)

AND

r=112

So the equation of circle is:

, (x12)2+(y14)2=(112)2

x2x+14+y212y+116=1144

x2+y2x12y1136=0

36x2+36y236x18y11=0

Question:4 Find the equation of the circle with

centre (1,1) and radius 2

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

So Given Here

(h,k)=(1,1)

AND r=2

So the equation of the circle is:

, (x1)2+(y1)2=(2)2

x22x+1+y22y+1=2

x2+y22x2y=0

Question:5 Find the equation of the circle with

centre (a,b) and radius a2b2

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

So Given Here

(h,k)=(a,b)

AND r=a2b2

So the equation of the circle is:

, (x(a))2+(y(b))2=(a2b2)2

x2+2ax+a2+y2+2by+b2=a2b2

x2+y2+2ax+2by+2b2=0

Question:6 Find the centre and radius of the circles.

(x+5)2+(y3)2=36

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

Given here

(x+5)2+(y3)2=36

Can also be written in the form

(x(5))2+(y3)2=62

So, from comparing, we can see that

r=6

Hence the Radius of the circle is 6.

Question:7 Find the centre and radius of the circles.

x2+y24x8y45=0

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

Given here

x2+y24x8y45=0

Can also be written in the form

(x2)2+(y4)2=(65)2

So, from comparing, we can see that

r=65

Hence the Radius of the circle is 65.

Question:8 Find the centre and radius of the circles.

x2+y28x+10y12=0

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

Given here

x2+y28x+10y12=0

Can also be written in the form

(x4)2+(y(5))2=(53)2

So, from comparing, we can see that

r=53

Hence the radius of the circle is 53.

Question:9 Find the centre and radius of the circles.

2x2+2y2x=0

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

Given here

2x2+2y2x=0

Can also be written in the form

(x14)2+(y0)2=(14)2

So, from comparing, we can see that

r=14

Hence Center of the circle is the (14,0)Radius of the circle is 14.

Question:10 Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x+y=16.

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

Given Here,

Condition 1: the circle passes through points (4,1) and (6,5)

(4h)2+(1k)2=r2

(6h)2+(5k)2=r2

Here,

(4h)2+(1k)2=(6h)2+(5k)2

(4h)2(6h)2+(1k)2(5k)2=0

(2)(102h)+(4)(62k)=0

20+4h24+8k=0

4h+8k=44

Now, Condition 2:centre is on the line 4x+y=16.

4h+k=16

From condition 1 and condition 2

h=3,k=4

Now lets substitute this value of h and k in condition 1 to find out r

(43)2+(14)2=r2

1+9=r2

r=10

So now, the Final Equation of the circle is

(x3)2+(y4)2=(10)2

x26x+9+y28y+16=10

x2+y26x8y+15=0

Question:11 Find the equation of the circle passing through the points (2,3) and (–1,1) and hose centre is on the line x3y11=0.

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

Given Here,

Condition 1: the circle passes through points (2,3) and (–1,1)

(2h)2+(3k)2=r2

(1h)2+(1k)2=r2

Here,

(2h)2+(3k)2=(1h)2+(1k)2

(2h)2(1h)2+(3k)2(1k)2=0

(3)(12h)+(2)(42k)=0

36h+84k=0

6h+4k=11

Now, Condition 2: centre is on the line.x3y11=0

h3k=11

From condition 1 and condition 2

h=72,k=52

Now let's substitute this value of h and k in condition 1 to find out r

(272)2+(3+52)2=r2

94+1214=r2

r2=1304

So now, the Final Equation of the circle is

(x72)2+(y+52)2=1304

x27x+494+y2+5y+254=1304

x2+y27x+5y564=0

x2+y27x+5y14=0

Question:12 Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

So let the circle be,

(xh)2+(yk)2=r2

Since it's radius is 5 and its centre lies on x-axis,

(xh)2+(y0)2=52

And Since it passes through the point (2,3).

(2h)2+(30)2=52

(2h)2=259

(2h)2=16

(2h)=4or(2h)=4

h=2or6

When h=2 ,The equation of the circle is :

(x(2))2+(y0)2=52

x2+4x+4+y2=25

x2+y2+4x21=0

When h=6 The equation of the circle is :

(x6)2+(y0)2=52

x212x+36+y2=25

x2+y212x+11=0

Question:13 Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Answer:

Let the equation of circle be,

(xh)2+(yk)2=r2

Now since this circle passes through (0,0)

(0h)2+(0k)2=r2

h2+k2=r2

Now, this circle makes an intercept of a and b on the coordinate axes.it means circle passes through the point (a,0) and (0,b)

So,

(ah)2+(0k)2=r2

a22ah+h2+k2=r2

a22ah=0

a(a2h)=0

a=0ora2h=0

Since a0soa2h=0

h=a2

Similarly,

(0h)2+(bk)2=r2

h2+b22bk+k2=r2

b22bk=0

b(b2k)=0

Since b is not equal to zero.

k=b2

So Final equation of the Circle ;

(xa2)2+(yb2)2=(a2)2+(b2)2

x2ax+a24+y2bx+b24=a24+b24

x2+y2axbx=0

Question:14 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Answer:

Let the equation of circle be :

(xh)2+(yk)2=r2

Now, since the centre of the circle is (2,2), our equation becomes

(x2)2+(y2)2=r2

Now, Since this passes through the point (4,5)

(42)2+(52)2=r2

4+9=r2

r2=13

Hence The Final equation of the circle becomes

(x2)2+(y2)2=13

x24x+4+y24y+4=13

x2+y24x4y5=0

Question:15 Does the point (–2.5, 3.5) lie inside, outside or on the circle x2+y2=25?

Answer:

Given, a circle

x2+y2=25

As we can see center of the circle is ( 0,0)

Now the distance between (0,0) and (–2.5, 3.5) is

d=(2.50)2+(3.50)2

d=6.25+12.25

d=18.54.3d=18.54.3<5

Since distance between the given point and center of the circle is less than radius of the circle, the point lie inside the circle.


Also read,


Topics covered in Chapter 10 Conic Section : Exercise 10.1

1. Introduction to Conic Sections: In this section, a brief idea about conic sections is provided. Various shapes related to conic sections such as circle, ellipse, parabola, hyperbola is dicussed having more emphasis upon circle..

2. Definition of a Circle: A circle is defined as the set of all points in a plane that are equidistant from a fixed point (called the centre). The fixed distance is called the radius.

3. Application-Based Questions: This involves the questions based on:

  • Identifying circle parameters.

  • Verifying if a point lies on a circle.

  • Finding missing values (like radius or coordinates of the centre).

  • Real-life applications involving the equation of a circle.


Class 11 Subject-Wise Solutions

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NCERT Solutions of Class 11 Subject Wise

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Frequently Asked Questions (FAQs)

1. What are the conics ?

The curves obtained as intersections of a plane with a double-napped right circular cone are called conics.

2. What is circle ?

The curve obtained from the set of all points in a plane that are equidistant from a fixed point in the plane is called a circle.

3. What is the radius of the circle ?

The distance between the centre to a point on the circle is called the radius of the circle.

4. Write the equation of the circle with centre at (0,0) and radius 3 unit.

Equation of the circle centre at (0,0) and radius 3  =>  x^2 + y^2 = 3^2

x^2 + y^2 = 9

5. Write the equation of the circle with centre at (1,2) and radius 2 unit.

Equation of the circle centre at (1,2) and radius 2  =>  (x-1)^2 + (y-2)^2 = 2^2

x^2 -2x + 1 + y^2 - 4y + 4= 4

x^2 -2x + y^2 - 4y + 1= 0

6. Find the centre and radius of the circle x^2 + y^2 + 4x + 6y – 12 = 0

Equation of the circle =>  x^2 + y^2 + 4x +  6y – 12 = 0

 (x+2)^2 + (y+3)^2 – 25 = 0

(x+2)^2 + (y+3)^2 = 5^2

Centre(-2,-3) and radius = 5.

Articles

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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