NCERT Solutions of class 11 maths exercise 11.1-Conic sections

NCERT Solutions of class 11 maths chapter 11 -Conic sections are the solutions of unsolved questions of the NCERT maths text book of class 11 chapter 11-Conic sections prescribed by CBSE a reputed school board of India.All NCERT Solutions of class 11 maths chapter 11 are solved by CBSE maths expert by a step by step way. All NCERT solutions of class 11 maths chapter 11 are explained here will help all maths students of class 11 and the candidates who are going to appear in competitive entrance exams like NDA,CDS,engineering entrance exams etc .

Download pdf of NCERT solutions class 11 chapter 11-Conic Section

PDF-NCERT solutions class 11 chapter 11-Conic Section

The chapter 11 of NCERT maths text book of class 11 is based on the –Conic Section, Conic Sections contains parabola,ellipse and hyperbola which are the planes carved out from the cone,it is that’s why chapter is named ‘Conic Section’. The questions are based on the equations of all the figure of parabola,ellipse and hyperbola.The chapter ‘Conic Section’ is easy so students can get an excellent score in maths by attempting the questions of chapter 11-Conic Sections in the maths question paper of class 11.

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Q1.FNCERT Solutions of class 11 maths chapter 11ind the equation of the circle with center (0,2) and radius 2.

Solution. The standard equation of the circle is

(x – h)² + (y – k)² = r²

Where (h,k) are the coordinates of the circle and r is the radius of the circle

It is given that center (h,k) = (0,2) and radius (r)=2

Therefore, equation of the circle is

(x – 0)² + (y – 2)² = (2)²

x² + y² + 4 – 4y = 4

x² + y² – 4y = 0

Q2.Find the equation of the circle with center (-2,3) and radius 4.

Ans. The equation of the circle with centre (h,k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that center (h,k) = (-2,3) and radius (r)= 4

Therefore, equation of the circle is

(x + 2)² + (y – 3)² = (4)²

x² + 4x² + 4 + y² – 6y + 9 = 16

x² + y² + 4x – 6y – 3 = 0

Q3. Find the equation of the circle with the centre (1/2, 1/4) and radius 1/12.

Ans. The equation of the circle with center (h,k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that center (h,k) = (-1/2,1/4) and radius (r)= 1/12

Therefore, the equation of the circle is

$\begin{pmatrix} x-\frac{1}{2} \end{pmatrix}^{2}+\begin{pmatrix} y-\frac{1}{4} \end{pmatrix}^{2}=\begin{pmatrix} \frac{1}{12} \end{pmatrix}^{2}$

$x^{2}-x+\frac{1}{4}+y^{2}-\frac{y}{2}+\frac{1}{16}=\frac{1}{144}$

$x^{2}-x+\frac{1}{4}+y^{2}-\frac{y}{2}+\frac{1}{16}-\frac{1}{144}=0$

144x² – 144x+ 36 + 144y² – 72y +9 – 1 = 0

144x² – 144x + 144y² – 72y + 44 = 0

36x² – 36x+ 36y² – 18y + 11 = 0

36x² + 36y² – 36x – 18y + 11 = 0

Q4. Find the equation of the circle with the center (1,1) and radius √2.

Ans. The equation of the circle with center (h,k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that center (h,k) = (1,1) and radius (r)=√2

Therefore, the equation of the circle is

(x – 1)² + (y – 1)² = (√2)²

x² – 2x +1 + y² – 2y + 1 = 2

x² + y² – 2x – 2y = 0

Q5. Find the equation of the circle with the center (-a,-b) and radius √(a² – b²).

Ans. The equation of the circle with center (h,k) and radius r is given as

(x – h)² + (y – k)² = r²

It is given that center (h,k) = (-a,-b) and radius (r)= √(a² – b²)

Therefore, the equation of the circle is

(x + a)² + (y + b)² = [√(a² – b²)]²

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

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

x² + y² + 2ax + 2by + 2b² = 0

Q6. Find the circle and radius of the circle (x + 5)² + (y – 3)² = 36

Ans. The equation of the given circle is

(x + 5)² + (y – 3)² = 36

{x -(- 5)}² + (y – 3)² = 6²

Which is of the form (x – h)² + (y – k)² = r², where h = -5, k = 3, and r = 3.

Thus, the center of given circle is (-5,3) while its radius is 6.

Q7. Find the circle and radius of the circle x² + y² – 4x – 8y – 45 = 0

Ans. The equation of the circle is

x² + y² – 4x – 8y – 45 = 0

(x² – 4x)+ (y² – 8y) = 45

{x² – 2(x)(2) + 2²} + {y² – 2(y)(4) + 4²} – 4 – 16 = 45

(x – 2)² + (y – 4)² = 65

(x – 2)² + (y – 4)² = (√65)²

Which is of the form (x – h)² + (y – k)² = r², where h = 2, k = 4, and r =√65

Thus, the center of the given circle is (2,4), while its radius is √65.

Q8. Find the center and radius of the circle is x² + y² – 8x +10y – 12 = 0

Ans. Standard equation of the circle is

(x – h)² + (y – k)² = r², where (h,k) are the co-ordinates of the centre of the circle.

The given equation of the circle is x² + y² – 8x +10y – 12 = 0

Rewritting the equation of the circle by applying the complete square method

x² – 8x + 4²- 4² + y²+10y +5²-5² – 12 = 0

(x – 4)² + (y + 5)² -16 – 25- 12 = 0

(x – 4)² + (y + 5)² = (√53)²

Comparing the standard equation of the circle

(x – h)² + (y – k)² = r²

h =4, k = -5 and r = √53

Therefore, the centre of the circle is (4,-5) and radius of the circle is √53

Q9. Find the centre and radius of the circle 2x² + 2y² -x = 0

Ans. The given equation of the circle is

2x² + 2y² -x = 0

Rearranging the equation by the method of complete square method

2x² -x + 2y² = 0

Dividing the equation by 2

x² – x/2 + y² = 0

Adding and subtracting the equation by 1/4

x² -x/2 + 1/4 – 1/4 + y² = 0

(x – 1/2)² + (y – 0)² = (1/2)²

(x – 1/2)² + (y – 0)² = (0.5)²

Now, comparing the equation with standard equation of the circle

(x – h)² + (y – k)² = r², where (h,k) are the co-ordinates of the centre of the circle.

Therefore co-ordinates of the centre are (1/2,0) and radius (0.5)

Q10.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

Ans. Let the equation of the required circle be (x – h)² + (y – k)² = r²

Since the circle passes through (4,1) and (6,5)

(4 – h)² + (1 – k)² = r²….(i)

(6 – h)² + (5 – k)² = r²….(ii)

Since the centre (h, k) of the circle lies on line 4x + y = 16

4h + k = 16….(iii)

From equations (i) and (ii), we obtain

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

16 – 8h +h² +1 – 2k +k² = 36 – 12h +h² +25 – 10k +k²

16 – 8h +1 – 2k = 36 – 12h +25 – 10k

4h + 8k = 44

h + 2k = 11….(iv)

On solving equations (iii) and (iv), we obtain h = 3 and k = 4

On substituting the values of h and k in equation (i), we obtain

(4 – 3)² + (1 – 4)² = r²

(1)² + (- 3)² = r²

1 + 9 = r²

r² = 10

r = √10

Thus, the equation of the required circle is

(x – 3)² + (y – 4)² = (√10)²

x² – 6x + 9 + y² – 8y + 16 = 0

x² + y² – 6x – 8y + 15 = 0

Q11.Find the equation of the circle passing through the points (2,3) and (-1,1) and whose centre is on the line x – 3y – 11 = 0

Ans. Let the equation of the required circle be (x – h)² + (y – k)² = r²

Since the circle passes thrpugh points (2,3) and (-1,1)

(2 – h)² + (3 – k)² = r²….(i)

(-1 – h)² + (1 – k)² = r²….(ii)

Since the centre (h,k) of the circle lies on line x – 3y – 11 = 0

h – 3k = 11….(iii)

From equations (i) and (ii), we obtain

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

4 – 4h + h² + 9 – 6k + k² = 1 + 2h + h² + 1 – 2k + k²

4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

6h + 4k = 11….(iv)

On solving equations (iii) and (iv), we obtain h = 7/2 and k = -5/2

On substituting the values of h and k in equation (i), we obtain

(2 – 7/2)² + (3 + 5/2)² = r²

[(4 – 7)/2)]² + [(6 + 5)/2)]² = r²

(-3/2)² + (11/2)² = r²

9/4 + 121/4 = r²

130/4 = r²

The equation of the required circle is

(x – 7/2)² + (y + 5/2)² = 130/4

[(2x – 7)/2)]² + [(2x + 5)/2)]² = 130/4

4x² – 28x + 49 +4y² + 20y +25 = 130

4x² +4y² – 28x + 20y – 56 = 0

4(x² +y² – 7x + 5y – 14) = 0

x² +y² – 7x + 5y – 14 = 0

∴ The equation of the required circle is x² +y² – 7x + 5y – 14 = 0

Q12.Find the equation of the circle with radius 5 cm whose centre lies on x-axis and passes through the point (2,3)

Ans. The standard equation of the circle is

(x – h)² + (y – k)² = r² where (h,k) are coordinates of the centre and r is the radius of the circle

We know that the radius of the circle is 5 and its centre lies on the x-axis, so coodinates of the centre are (h,k) =(h,0) and r = 5

So now, the equation of the circle is (x – h)² + y ² = 25

It is given that the circle passes through the point (2,3) so the point will satisfy the equation of the circle.

(2 – h)² + 3² = 25

(2 – h)² = 25 – 9

(2 – h)² = 16

2 – h = ± 4

h = 2± 4

h = 6 or -2

The required equation of circle is

(x-6)² +y² =5² or (x + 2)² +k² = 5²

x² + 36 – 12x + y² = 25, x² + 2² +4x +k² = 25

x² – 12x + y² +11=0, x² +4x +k² -21 =0

Q13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Ans.The standard equation of the circle is (x – h)² + (y – k)² = r² where (h,k) are coordinates of the centre and r is the radius of the circle

We are given that circle is passing through (0,0) and making intercepts a and b on the coordinate axes means circle is passing through the point (a,0) on x-axis and (0,b) on y-axis.

substituting (0,0) , (a,0) and (b,0) in the equation of the circle

h² + k² = r²…….(i)

(a -h)² + k²=r²…….(ii)

h² + (b -k)²=r² …….(ii)

from equation (i) and (ii), we have

(a -h)² + k²= h² + k²

a² + h² -2ah + k² = h² + k²

a² – 2ah = 0 ⇒ a(a – 2h) = 0 ⇒a = 0 and a = 2h

a ≠0, so h = a/2

similarly from equation (iii) and (i), we have

(b -k)² + h²= h² + k²

b² + k² -2bk + h² = h² + k²

b² – 2bk = 0 ⇒ b(b – 2k) = 0 ⇒b = 0 and b = 2k

b ≠0, so k= b/2

Now, sbstituting the value of (h,k) =(a/2, b/2) in equation (i)

a²/4 + b²/4 = r²

$\boldsymbol{r=\frac{\sqrt{a^{2}+b^{2}}}{2}}$

Putting the value of (h,k) and r in the equation of circle

$\boldsymbol{\left ( x-\frac{a}{2} \right )^{2}+\left ( y-\frac{b}{2} \right )^{2}=\frac{a^{2}-b^{2}}{4}}$

$\boldsymbol{\frac{}{}x^{2}+\frac{a^{2}}{4}-ax +y^{2}+\frac{b^{2}}{4}-by=\frac{a^{2}+b^{2}}{4}}$

4x²+ 4y² – 4ax – 4by = 0

x²+ y² – ax – by = 0 is the required equation

Q14.Find the equation of a circle with centre (2,2) and passes through point (4,5).

Ans. The standard equation of the circle is (x – h)² + (y – k)² = r² where (h,k) are coordinates of the centre and r is the radius of the circle

We are given that circle is passing through (4,5) and the coordinates(h,k) of the centre are (2,2)

So,we have r² = (4 -2)² + (5 -2)²= 4 + 9 = 13

r = √13

The equation of the circle is

(x -2)² + (y -2)² =13

x ² + 4 -4x + y² + 4 – 4y = 13

x ² + y² -4x – 4y +8 -13 =0

x ² + y² -4x – 4y -5 =0

Q15.Does the point(-2.5,3.5) lie inside,outside or on the circle x² + y² = 25 ?

Ans. The given equation of the circle is x² + y² = 25

Comparing the given equation with the standard equation of the circle

(x – h)² + (y – k)² = r² where (h,k) are coordinates of the centre and r is the radius of the circle

We get h =0,k =0 and r = 5

Extract of the exercise 11.1 of the chapter 11-Conic Section

The centre of the circle is (0,0) and radius 5 indicates that the point (-2.5,3.5) lies inside the circle since the distance of the point from the centre(0,0) is less than 5 , the radius of the circle.

The circle is the most common geometrical figure in a range of all conic sections. The circle can be driven by slicing a plane parallel to its base and perpendicular to the axis of the cone.

The standard equation of the circle

The standard equation of the circle is

(x – h)² + (y – k)² = r²

Where (h,k) are the coordinates of the circle and r is the radius of the circle

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals