NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1: All exercise 11.1 questions are answered in the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections. The goal of practicing NCERT Solutions is to improve your final test result. To obtain the solutions to all of the exercises in all of the chapters, download NCERT Class 11 maths solutions.
A conic section is a chapter that covers various cone sections. In its first exercise, it covers a wide range of themes. The following subjects are covered in Exercise 11.1 of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections:
- Introduction
- Sections of a Cone
- Circle, ellipse, parabola, and hyperbola
- Degenerated conic sections
- Circle
Download NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1 Free PDF
NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1
Access the NCERT Solutions for Class 11 Maths Chapter 11 – Exercise 11.1
In each of the following Exercises 1 to 5, find the equation of the circle with
1. center (0, 2) and radius 2.
Solution:
Given:
center (0, 2) and radius 2.
Let us consider the equation of a circle with center (h, k), and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, center (h, k) = (0, 2) and radius (r) = 2
The equation of the circle is
(x – 0)2 + (y – 2)2 = 22
x2 + y2 + 4 – 4y = 4
x2 + y2 – 4y = 0
∴ The equation of the circle is x2 + y2 – 4y = 0
2. center (–2, 3) and radius 4.
Solution:
Given:
center (-2, 3) and radius 4
Let us consider the equation of a circle with center (h, k), and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, center (h, k) = (-2, 3) and radius (r) = 4
The equation of the circle is
(x + 2)2 + (y – 3)2 = (4)2
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
3. center (1/2, 1/4) and radius (1/12).
Solution:
Given:
center (1/2, 1/4) and radius 1/12
Let us consider the equation of a circle with center (h, k), and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, center (h, k) = (1/2, 1/4) and radius (r) = 1/12
The equation of the circle is
(x – 1/2)2 + (y – 1/4)2 = (1/12)2
x2 – x + ¼ + y2 – y/2 + 1/16 = 1/144
x2 – x + ¼ + y2 – y/2 + 1/16 = 1/144
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
∴ The equation of the circle is 36x2 + 36y2 – 36x – 18y + 11= 0
4. center (1, 1) and radius √2
Solution:
Given:
center (1, 1) and radius √2
Let us consider the equation of a circle with center (h, k), and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, center (h, k) = (1, 1) and radius (r) = √2
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
∴ The equation of the circle is x2 + y2 – 2x -2y = 0
5. center (–a, –b) and radius √(a2 – b2)
Solution:
Given:
center (-a, -b) and radius √(a2 – b2)
Let us consider the equation of a circle with center (h, k), and
Radius r is given as (x – h)2 + (y – k)2 = r2
So, center (h, k) = (-a, -b) and radius (r) = √(a2 – b2)
The equation of the circle is
(x + a)2 + (y + b)2 = (√(a2 – b2)2)
x2 + 2ax + a2 + y2 + 2by + b2 = a2 – b2
x2 + y2 +2ax + 2by + 2b2 = 0
∴ The equation of the circle is x2 + y2 +2ax + 2by + 2b2 = 0
In each of the following Exercises 6 to 9, find the center and radius of the circles.
6. (x + 5)2 + (y – 3)2 = 36
Solution:
Given:
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 = 6
∴ The center of the given circle is (-5, 3), and its radius is 6.
7. x2 + y2 – 4x – 8y – 45 = 0
Solution:
Given:
The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0
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 in the form (x-h)2 +(y-k)2 = r2]
Where h = 2, K = 4 and r = √65
∴ The center of the given circle is (2, 4), and its radius is √65.
8. x2 + y2 – 8x + 10y – 12 = 0
Solution:
Given:
The equation of the given circle is x2 + y2 -8x + 10y -12 = 0
x2 + y2 – 8x + 10y – 12 = 0
(x2 – 8x) + (y2 + 10y) = 12
(x2 – 2(x) (4) + 42) + (y2 – 2(y) (5) + 52) – 16 – 25 = 12
(x – 4)2 + (y + 5)2 = 53
(x – 4)2 + (y – (-5))2 = (√53)2 [which is in the form (x-h)2 +(y-k)2 = r2]
Where h = 4, K= -5 and r = √53
∴ The center of the given circle is (4, -5), and its radius is √53.
9. 2x2 + 2y2 – x = 0
Solution:
The equation of the given circle is 2x2 + 2y2 –x = 0
2x2 + 2y2 –x = 0
(2x2 + x) + 2y2 = 0
(x2 – 2 (x) (1/4) + (1/4)2) + y2 – (1/4)2 = 0
(x – 1/4)2 + (y – 0)2 = (1/4)2 [which is in the form (x-h)2 +(y-k)2 = r2]
Where, h = ¼, K = 0, and r = ¼
∴ The center of the given circle is (1/4, 0), and its radius is 1/4.
10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose center is on line 4x + y = 16.
Solution:
Let us consider the equation of the required circle be (x – h)2+ (y – k)2 = r2
We know that the circle passes through points (4,1) and (6,5).
So,
(4 – h)2 + (1 – k)2 = r2 ……………..(1)
(6– h)2+ (5 – k)2 = r2 ………………(2)
The center (h, k) of the circle lies on line 4x + y = 16.
4h + k =16………………… (3)
From the equation (1) and (2), we obtain
(4 – h)2+ (1 – k)2 =(6 – h)2 + (5 – k)2
16 – 8h + h2 +1 -2k +k2 = 36 -12h +h2+15 – 10k + k2
16 – 8h +1 -2k + 12h -25 -10k
4h +8k = 44
h + 2k =11……………. (4)
On solving equations (3) and (4), we obtain h=3 and k= 4.
On substituting the values of h and k in equation (1), we obtain
(4 – 3)2+ (1 – 4)2 = r2
(1)2 + (-3)2 = r2
1+9 = r2
r = √10
So now, (x – 3)2 + (y – 4)2 = (√10)2
x2 – 6x + 9 + y2 – 8y + 16 =10
x2 + y2 – 6x – 8y + 15 = 0
∴ The equation of the required circle is x2 + y2 – 6x – 8y + 15 = 0
11. Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose center is on the line x – 3y – 11 = 0.
Solution:
Let us consider the equation of the required circle be (x – h)2 + (y – k)2 = r2
We know that the circle passes through points (2,3) and (-1,1).
(2 – h)2+ (3 – k)2 =r2 ……………..(1)
(-1 – h)2+ (1– k)2 =r2 ………………(2)
The center (h, k) of the circle lies on line x – 3y – 11= 0.
h – 3k =11………………… (3)
From the equation (1) and (2), 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……………. (4)
Now, let us multiply equation (3) by 6 and subtract it from equation (4) to get
6h+ 4k – 6(h-3k) = 11 – 66
6h + 4k – 6h + 18k = 11 – 66
22 k = – 55
K = -5/2
Substitute this value of K in equation (4) to get
6h + 4(-5/2) = 11
6h – 10 = 11
6h = 21
h = 21/6
h = 7/2
We obtain h = 7/2and k = -5/2
On substituting the values of h and k in equation (1), we get
(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 + [(2y+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
12. Find the equation of the circle with radius 5 whose center lies on the x-axis and passes through the point (2, 3).
Solution:
Let us consider the equation of the required circle be (x – h)2+ (y – k)2 = r2
We know that the radius of the circle is 5 and its center lies on the x-axis, k = 0 and r = 5.
So now, the equation of the circle is (x – h)2 + y2 = 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 = ± √16 = ± 4
If 2-h = 4, then h = -2
If 2-h = -4, then h = 6
Then, when h = -2, the equation of the circle becomes
(x + 2)2 + y2 = 25
x2 + 12x + 36 + y2 = 25
x2 + y2 + 4x – 21 = 0
When h = 6, the equation of the circle becomes
(x – 6)2 + y2 = 25
x2 -12x + 36 + y2 = 25
x2 + y2 -12x + 11 = 0
∴ The equation of the required circle is x2 + y2 + 4x – 21 = 0 and x2 + y2 -12x + 11 = 0
13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.
Solution:
Let us consider the equation of the required circle be (x – h)2+ (y – k)2 =r2
We know that the circle passes through (0, 0).
So, (0 – h)2+ (0 – k)2 = r2
h2 + k2 = r2
Now, the equation of the circle is (x – h)2 + (y – k)2 = h2 + k2.
It is given that the circle intercepts a and b on the coordinate axes.
i.e., the circle passes through points (a, 0) and (0, b).
So, (a – h)2+ (0 – k)2 =h2 +k2……………..(1)
(0 – h)2+ (b– k)2 =h2 +k2………………(2)
From equation (1), we obtain
a2 – 2ah + h2 +k2 = h2 +k2
a2 – 2ah = 0
a(a – 2h) =0
a = 0 or (a -2h) = 0
However, a ≠ 0; hence, (a -2h) = 0
h = a/2
From equation (2), we obtain
h2 – 2bk + k2 + b2= h2 +k2
b2 – 2bk = 0
b(b– 2k) = 0
b= 0 or (b-2k) =0
However, a ≠ 0; hence, (b -2k) = 0
k =b/2
So, the equation is
(x – a/2)2 + (y – b/2)2 = (a/2)2 + (b/2)2
[(2x-a)/2]2 + [(2y-b)/2]2 = (a2 + b2)/4
4x2 – 4ax + a2 +4y2 – 4by + b2 = a2 + b2
4x2 + 4y2 -4ax – 4by = 0
4(x2 +y2 -7x + 5y – 14) = 0
x2 + y2 – ax – by = 0
∴ The equation of the required circle is x2 + y2 – ax – by = 0
14. Find the equation of a circle with center (2,2) and passes through the point (4,5).
Solution:
Given:
The center of the circle is given as (h, k) = (2,2).
We know that the circle passes through the point (4,5), and the radius (r) of the circle is the distance between points (2,2) and (4,5).
r = √[(2-4)2 + (2-5)2]
= √[(-2)2 + (-3)2]
= √[4+9]
= √13
The equation of the circle is given as
(x– h)2+ (y – k)2 = r2
(x –h)2 + (y – k)2 = (√13)2
(x –2)2 + (y – 2)2 = (√13)2
x2 – 4x + 4 + y2 – 4y + 4 = 13
x2 + y2 – 4x – 4y = 5
∴ The equation of the required circle is x2 + y2 – 4x – 4y = 5
15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Solution:
Given:
The equation of the given circle is x2 +y2 = 25
x2 + y2 = 25
(x – 0)2 + (y – 0)2 = 52 [which is in the form (x – h)2 + (y – k)2 = r2]
Where, h = 0, k = 0 and r = 5
So, the distance between the point (-2.5, 3.5) and the center (0,0) is
= √[(-2.5 – 0)2 + (-3.5 – 0)2]
= √(6.25 + 12.25)
= √18.5
= 4.3 [which is < 5]
Since the distance between the point (-2.5, -3.5) and the center (0, 0) of the circle is less than the radius of the circle, the point (-2.5, -3.5) lies inside the circle.
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FAQs on NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1
Here are the most frequently asked questions related to NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 by the students.
What is Chapter 11 of NCERT Maths of Class 11?
Chapter 11 of NCERT Maths of Class 11 is Conic Sections.
Where can I get Class 11 Maths NCERT Solutions for Chapter 11 Exercise 11.1?
You can get the NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1 from the pdf given in the article above.
How many exercises are there in Chapter 11 of Maths NCERT of Class 11?
There are a total of 4 exercises and one miscellaneous exercise in chapter 11 of Maths NCERT of Class 11.
What are some of the benefits of using NCERT Math Solutions?
Maths NCERT Solutions has the advantage of allowing you to learn the proper solution while studying the problem and using it to guide you in the right direction. You will be able to save time as a result of it.
What is exercise 11.1 of the NCERT Maths of Class 11 Chapter 11 based on?
The following subjects are covered in Exercise 11.1 of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections:
Introduction
Sections of a Cone
Circle, ellipse, parabola and hyperbola
Degenerated conic sections
Circle
What are NCERT Maths Solutions for Class 11?
Answers have been produced by NCERT specialists based on the most recent CBSE guidelines. All chapters from the CBSE 11th-grade math textbooks are included.