RD Sharma Solutions Class 11 Maths Chapter 24 – The Circle: RD Sharma Solutions Class 11 Maths Chapter 24 The Circle assists the students in getting to know about the exam pattern so as to achieve higher marks in exams. The solutions are developed with shortcut techniques that make learning easy. The students can attain good marks in the final exams by practicing these solutions regularly from RD Sharma Solutions Class 11 Maths.
Download RD Sharma Solutions Class 11 Maths Chapter 24 – The Circle PDF
RD Sharma Solutions Class 11 Maths Chapter 24
Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 24 PDF
| RD Sharma Class 11 Chapter 24 Exercise 24a |
| RD Sharma Class 11 Chapter 24 Exercise 24b |
| RD Sharma Class 11 Chapter 24 Exercise 24c |
Detailed Exercise-wise Explanation with a Listing of Important Topics in the Exercise
RD Sharma Solutions Class 11 Maths Chapter 24 are solved by experienced maths faculty with easy methods. These solutions include well-researched information about each exercise so that the students can easily understand all concepts.
These solutions enable the students to get confident about their answers by practising RD Sharma’s class 11 solutions Maths Chapter 24 The Circle on a regular basis. The students can refer to exercise-wise solutions provided with short keynotes that can boost their exam preparations.
RD Sharma Class 11 Chapter 24 Exercise 24a
This exercise includes finding the equation of a circle when its center & radius are known & finding the center and radius of a given circle. RD Sharma class 11 chapter 24 exercise 24a assists the students to get excellent marks in the exams.
The solutions are prepared by the maths expert faculty in a step-by-step manner to assist the students in understanding the topics clearly. The students must practice exercise-wise problems regularly in order to attain better marks in the exams.
RD Sharma Class 11 Chapter 24 Exercise 24b
This exercise includes problems related to the topics of the general equation of a circle. The students can strengthen their knowledge & build a strong command over the topics of this chapter with the help of these solutions.
The Mathematics experts have solved RD Sharma Class 11 Chapter 24 Exercise 24b in an easy & understandable language.
RD Sharma Class 11 Chapter 24 Exercise 24c
This exercise includes problems based on the topics of the diameter form of a circle. The students can refer to this exercise solution as a reference guide to enhance their conceptual knowledge & understand the different methods to solve the problems.
RD Sharma class 11 chapter 24 exercise 24c is created by the Maths expert faculty to improve students’ confidence to get good marks.
Access RD Sharma Solutions Class 11 Maths Chapter 24
EXERCISE 24.1 PAGE NO: 24.21
1. Find the equation of the circle with:
(i) Centre (-2, 3) and radius 4.
(ii) Centre (a, b) and radius
.
(iii) Centre (0, – 1) and radius 1.
(iv) Centre (a cos α, a sin α) and radius a.
(v) Centre (a, a) and radius √2 a.
Solution:
(i) Centre (-2, 3) and radius 4.
Given:
The radius is 4, and the center (-2, 3)
By using the formula,
The equation of the circle with center (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 = r2
Where, p = -2, q = 3, r = 4
Now, by substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – (-2))2 + (y – 3)2 = 42
(x + 2)2 + (y – 3)2 = 16
x2 + 4x + 4 + y2 – 6y + 9 = 16
x2 + y2 + 4x – 6y – 3 = 0
∴ The equation of the circle is x2 + y2 + 4x – 6y – 3 = 0
(ii) Centre (a, b) and radius
.
Given:
The radius is
and the centre (a, b)
By using the formula,
The equation of the circle with center (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 = r2
Where, p = a, q = b, r =
Now, by substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – a)2 + (y – b)2 =
x2 – 2ax + a2 + y2 – 2by + b2 = a2 + b2
x2 + y2 – 2ax – 2by = 0
∴ The equation of the circle is x2 + y2 – 2ax – 2by = 0
(iii) Centre (0, -1) and radius 1.
Given:
The radius is 1, and the center (0, -1)
By using the formula,
The equation of the circle with center (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 = r2
Where, p = 0, q = -1, r = 1
Now, by substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – 0)2 + (y – (-1))2 = 12
(x – 0)2 + (y + 1)2 = 1
x2 + y2 + 2y + 1 = 1
x2 + y2 + 2y = 0
∴ The equation of the circle is x2 + y2 + 2y = 0.
(iv) Centre (a cos α, a sin α) and radius a.
Given:
The radius is ‘a’, and the center (a cos α, a sin α)
By using the formula,
The equation of the circle with center (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 = r2
Where, p = a cos α, q = a sin α, r = a
Now, by substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – a cosα)2 + (y – a sinα)2 = a2
x2 – (2acosα)x + a2cos2α + y2 – (2asinα)y + a2sin2α = a2
We know that sin2θ + cos2θ = 1
So,
x2 – (2acosα)x + y2 – 2asinαy + a2 = a2
x2 + y2 – (2acosα)x – (2asinα)y = 0
∴ The equation of the circle is x2 + y2 – (2acosα) x – (2asinα) y = 0.
(v) Centre (a, a) and radius √2 a.
Given:
The radius is √2 a, and the center (a, a)
By using the formula,
The equation of the circle with center (p, q) and radius ‘r’ is (x – p)2 + (y – q)2 = r2
Where, p = a, q = a, r = √2 a
Now, by substituting the values in the above equation, we get
(x – p)2 + (y – q)2 = r2
(x – a)2 + (y – a)2 = (√2 a)2
x2 – 2ax + a2 + y2 – 2ay + a2 = 2a2
x2 + y2 – 2ax – 2ay = 0
∴ The equation of the circle is x2 + y2 – 2ax – 2ay = 0.
2. Find the center and radius of each of the following circles:
(i) (x – 1)2 + y2 = 4
(ii) (x + 5)2 + (y + 1)2 = 9
(iii) x2 + y2 – 4x + 6y = 5
(iv) x2 + y2 – x + 2y – 3 = 0
Solution:
(i) (x – 1)2 + y2 = 4
Given:
The equation (x – 1)2 + y2 = 4
We need to find the center and the radius.
By using the standard equation formula,
(x – a)2 + (y – b)2 = r2 …. (1)
Now, let us convert the given circle’s equation into the standard form.
(x – 1)2 + y2 = 4
(x – 1)2 + (y – 0)2 = 22 ….. (2)
By comparing equation (2) with (1), we get
Centre = (1, 0) and radius = 2
∴ The center of the circle is (1, 0), and the radius is 2.
(ii) (x + 5)2 + (y + 1)2 = 9
Given:
The equation (x + 5)2 + (y + 1)2 = 9
We need to find the center and the radius.
By using the standard equation formula,
(x – a)2 + (y – b)2 = r2 …. (1)
Now let us convert the given circle’s equation into the standard form.
(x + 5)2 + (y + 1)2 = 9
(x – (-5))2 + (y – ( – 1))2 = 32 …. (2)
By comparing equation (2) with (1), we get
Centre = (-5, -1) and radius = 3
∴ The center of the circle is (-5, -1), and the radius is 3.
(iii) x2 + y2 – 4x + 6y = 5
Given:
The equation x2 + y2 – 4x + 6y = 5
We need to find the center and the radius.
By using the standard equation formula,
(x – a)2 + (y – b)2 = r2 …. (1)
Now let us convert the given circle’s equation into the standard form.
x2 + y2 – 4x + 6y = 5
(x2 – 4x + 4) + (y2 + 6y + 9) = 5 + 4 + 9
(x – 2)2 + (y + 3)2 = 18
(x – 2)2 + (y – (-3))2 = (3√2)2 … (2)
By comparing equation (2) with (1), we get
Centre = (2, -3) and radius = 3√2
∴ The center of the circle is (2, -3), and the radius is 3√2.
(iv) x2 + y2 – x + 2y – 3 = 0
Given:
The equation x2 + y2 – x + 2y – 3 = 0
We need to find the center and the radius.
By using the standard equation formula,
(x – a)2 + (y – b)2 = r2 …. (1)
Now let us convert the given circle’s equation into the standard form.
x2 + y2 – x + 2y – 3 = 0
(x2 – x + ¼) + (y2 + 2y + 1) – 3 – ¼ – 1 = 0
(x – ½)2 + (y + 1)2 = 17/4 …. (2)
By comparing equation (2) with (1), we get
Centre = (½, – 1) and radius = √17/2
∴ The center of the circle is (½, -1), and the radius is √17/2.
3. Find the equation of the circle whose center is (1, 2) and which passes through the point (4, 6).
Solution:
Given:
The Centre is (1, 2) and passes through the point (4, 6).
Where, p = 1, q = 2
We need to find the equation of the circle.
By using the formula,
(x – p)2 + (y – q)2 = r2
(x – 1)2 + (y – 2)2 = r2
It passes through the point (4, 6)
(4 – 1)2 + (6 – 2)2 = r2
32 + 42 = r2
9 + 16 = r2
25 = r2
r = √25
= 5
So r = 5 units
We know that the equation of the circle with center (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2
By substituting the values in the above equation, we get
(x – 1)2 + (y – 2)2 = 52
x2 – 2x + 1 + y2 – 4y + 4 = 25
x2 + y2 – 2x – 4y – 20 = 0.
∴ The equation of the circle is x2 + y2 – 2x – 4y – 20 = 0.
Important Topics Class 11 Maths Chapter 24 – The Circle
RD Sharma Solutions Class 11 Maths Chapter 24 covers some important concepts that are listed below:
- Standard equation of a circle.
- General equations of a circle.
- Diameter form of a circle.
RD Sharma Solutions Class 11 Maths Chapter 24 – The Circle
RD Sharma Solutions Class 11 Maths Chapter 24 enables the students to learn about the circle & also find the equation of any circle whose center & radius are given. These solutions are provided in the most comprehensible & easy manner. The solutions are solved using shortcut techniques to enable the students to grasp the topics quickly & easily.
The students must go refer to RD Sharma’s class 11 Solutions Maths chapter 24 The Circle to acquire better marks in final exams. If students can practice exercise-wise solutions on a regular basis, they can definitely attain good marks in the exams. These solutions are accurate and reliable & solved as per the CBSE guidelines.
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