R S Aggarwal Solutions for Class 11 Maths Chapter 21 Circles Exercise 21B

R S Aggarwal Solutions Class 11 Maths are given here for Chapter 21 – Circles exercise 21B. We are providing R S Aggarwal Chapter 21 solutions to help students go through and practice several solved problems. In this exercise, students will study about general equation of a circle.

Important Note:

The general equation of a circle is of the form x2 + y2 + 2gx + 2fy + c = 0.

Where (-g, -f) is the centre and radius = sqrt(g2 + f2 – c)

Also, every equation of the form x2 + y2 + 2gx + 2fy + c = 0 represents a circle if (g2 + f2 – c) > 0.

Find detailed solutions for chapter 21 exercise 21B below.

Download PDF of R S Aggarwal Solutions for Class 11 Maths Chapter 21 Circles Exercise 21B

 

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 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21
 rs aggarwal solution class 11 maths chapter 21

 

Access Answers to Maths R S Aggarwal Class 11 Chapter 21 Circles Exercise 21B Page number 730

Exercise 21B Page No: 730

Question 1: Show that the equation x2 + y2 – 4x + 6y – 5 = 0 represents a circle. Find its centre and radius.

Solution:

Given equation is x2 + y2 – 4x + 6y – 5 = 0

The general equation of a circle is as follows:

x2 + y2 + 2gx + 2fy + c = 0

Where g, f and c are constants

With

Centre: (-g, -f)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 1

On comparing given equation with general form of circle, we have

2g = – 4 ⇒ g = -2

2f = 6 ⇒ f = 3 and

c = – 5

Centre: (-g, -f) = (2, -3)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 1 solution

Question 2: Show that the equation x2 + y2 + x – y = 0 represents a circle. Find its centre and radius.

Solution:

Given equation is x2 + y2 + x – y = 0

The general equation of a circle is as follows:

x2 + y2 + 2gx + 2fy + c = 0

Where g, f and c are constants

With

Centre: (-g, -f)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 2

On comparing given equation with general form of circle, we have

2g = 1 ⇒ g = 1/2

2f = – 1 ⇒ f = -1/2 and

c = 0

Now,

Centre: (-g, -f) = (-1/2, 1/2)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 2 solution

Question 3: Show that the equation 3x2 + 3y2 + 6x – 4y – 1 = 0 represents a circle. Find its centre and radius.

Solution:

Given equation is 3x2 + 3y2 + 6x – 4y – 1 = 0

Or x2 + y2 + 2x – 4/3 y – 1/3 = 0

The general equation of a circle is as follows:

x2 + y2 + 2gx + 2fy + c = 0

Where g, f and c are constants.

With

Centre: (-g, -f)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 3

On comparing given equation with general form of circle, we have

2g = 2 ⇒ g = 1

2f = -4/3 ⇒ f = -2/3 and

c = -1/3

Now,

Centre (-g, -f) = (-1, 2/3)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 3 solution

Question 4: Show that the equation x2 + y2 + 2x + 10y + 26 = 0 represents a point circle. Also, find its centre.

Solution:

Given equation is x2 + y2 + 2x + 10y + 26 = 0

The general equation of a circle:

x2 + y2 + 2gx + 2fy + c = 0

where c, g, f are constants.

On comparing given equation with general equation of circle, we have

2g = 2 ⇒ g = 1

2f = 10 ⇒ f = 5 and

c = 26

Now,

Centre ( – g, – f) = ( – 1, – 5).

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 4

Since radius is zero, Thus it is a point circle with radius zero.

Question 5: Show that the equation x2 + y2 – 3x + 3y + 10 = 0 does not represent a circle.

Solution:

Given equation is x2 + y2 – 3x + 3y + 10 = 0

We know that, any equation with negative radius(complex number) does not represent a circle.

Find radius of given equation:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 5

Radius is a complex number. Therefore, given equation does not represent a circle.

Question 6: Find the equation of the circle passing through the points

(i) (0, 0), (5, 0) and (3, 3)

(ii) (1, 2), (3, – 4) and (5, – 6)

(iii) (20, 3), (19, 8) and (2, – 9)

Also, find the centre and radius in each case.

Solution:

Before we start solving listed problems, students are advised to keep below information in mind.

The general equation of a circle is as follows:

x2 + y2 + 2gx + 2fy + c = 0

Where g, f and c are constants.

With

Centre: (-g, -f)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6

(i) (0, 0), (5, 0) and (3, 3)

The Circle equation is:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 1

Let us apply Laplace Expansion to solve this problem:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 1 solution

On comparing above equation with the general form of circle, we get

2g = -5 ⇒ g = -2.5

2f = -1 ⇒ f = -0.5

c = 0

Now,

centre = (2.5, 0.5)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 1 solutions

(ii) (1, 2), (3, – 4) and (5, – 6)

The Circle equation is:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part ii solution

Let us apply Laplace Expansion to solve this problem:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 2 solution

On comparing above equation with the general form of circle, we get

2g = -22 ⇒ g = -11

2f = -4 ⇒ f = -2

c = -25

Now,

Centre = (11, 2)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part ii solutions

(iii) (20, 3), (19, 8) and (2, – 9)

The Circle equation is:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part 3

Let us apply Laplace Expansion to solve this problem:

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part iii solution

On comparing above equation with the general form of circle, we get

2g = -14 ⇒ g = -7

2f = -6 ⇒ f = -3

c = -111

Now,

Centre = (7, 3)

R S Aggarwal Solution Class 11 chapter 21 ex 21B question 6 part  3 solution

Question 7: Find the equation of the circle which is circumscribed about the triangle whose vertices are A( – 2, 3), b(5, 2) and C(6, – 1). Find the centre and radius of this circle.

Solution:

Since circle is circumscribed about the triangle whose vertices are A( – 2, 3), B(5, 2) and C(6, – 1), which implies points A, B and C are lie on circumference of circle and satisfy its equation.

The general equation of a circle: (x – h)2 + (y – k) 2 = r2 …(i)

where (h, k) is the centre and r is the radius.

Putting A(-2, 3), B(5, 2) and C(6, -1) in above equation, we get

h2 + k2 + 4h – 6k + 13 = r2 ……….(ii)

h2 + k2 – 10h – 4k + 29 = r2 ………(iii)

h2 + k2 – 12h + 2k + 37 = r2 ………(iv)

Subtract (ii) from (iii)

– 14h + 2k + 16 = 0

or – 7h + k + 8 = 0 ….(v)

Subtract (ii) from (iv)

– 16h + 8k + 24 = 0

or -2h + k + 3 = 0 ……(vi)

Solving (v) and (vi), we have

h = 1

(vi) ⇒ -2 x 1 + k + 3 = 0

⇒ k= -1

Therefore,

Centre = (1, – 1)

And,

Equation (ii) ⇒ r = 5

[using values of h and k]

Thus, required equation of the circle is

(x – 1) 2 + (y + 1) 2 = 52

(x – 1) 2 + (y + 1) 2 = 25

Question 8: Find the equation of the circle concentric with the circle x2 + y2 + 4x + 6y + 11 = 0 and passing through the point P(5, 4).

Solution:

Since circles are concentric, which means circles have common centre and different radii.

Equation of given circle, x2 + y2 + 4x + 6y + 11 = 0

The concentric circle will have the equation

x2 + y2 + 4x + 6y + d = 0 ….(1)

As it passes through P(5, 4),

Put x = 5 and y = 4

52 + 42 + 20 + 24 + d= 0

25 + 16 + 20 + 24 + d = 0

d = – 85

Equation (1) ⇒ x2 + y2 + 4x + 6y – 85 = 0

Which is required equation.

Question 9: Show that the points A(1, 0), B(2, – 7), C(8, 1) and D(9, – 6) all lie on the same circle. Find the equation of this circle, its centre and radius.

Solution:

The general equation of a circle: (x – h)2 + (y – k) 2 = r2 …(i)

where (h, k) is the centre and r is the radius.

Consider points (1, 0), (2, – 7) and (8, 1) lie on the circle.

Putting (1, 0), (2, – 7) and (8, 1) in (i)

Putting (1, 0) ⇒ h2 + k2 + 1 – 2h = r2 ……(ii)

Putting (2, – 7) ⇒ h2 + k2 + 53 – 4h + 14k = r2 …….(iii)

Putting (8, 1) ⇒ (8 – h)2 + (1 – k) 2 = r2

h2 + k2 + 65 – 16h – 2k = r2 ………….(iv)

Subtract (ii) from (iii), we get

h – 7k – 26 = 0 ……(v)

Subtract (ii) from (iv), we get

7h + k – 32 = 0 ……(vi)

Solving (v) and (vi)

h = 5 and k = – 3

Equation (iv) ⇒ r = 25

[using h = 5 and k = – 3]

Therefore,

Centre (5, – 3)

Radius = 25

Check for (9, – 6):

To check if (9, – 6) lies on the circle,

(9 – 5)2 + ( – 6 + 3)2 = 52

25 = 25

Which is true.

Hence, all the points are lie on circle.


Access other exercise solutions of Class 11 Maths Chapter 21 Circles

Exercise 21A Solutions

R S Aggarwal Solutions for Class 11 Maths Chapter 21 Exercise 21B

Class 11 Maths Chapter 21 Circles Exercise 21B is based on general equation of a circle. Students can download R S Aggarwal chapter 21 solutions and practice more problems on circle.