**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 .

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**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

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

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

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

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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 |

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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

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