NCERT Solutions for Exercise 11.1 Class 11 Maths Chapter 11

Q: What are the conics ?

The curves obtained as intersections of a plane with a double-napped right circular cone are called conics.

Q: What is circle ?

The curve obtained from the set of all points in a plane that are equidistant from a fixed point in the plane is called a circle.

Q: What is the radius of the circle ?

The distance between the centre to a point on the circle is called the radius of the circle.

Q: Write the equation of the circle with centre at (0,0) and radius 3 unit.

Equation of the circle centre at (0,0) and radius 3 => x^2 + y^2 = 3^2 x^2 + y^2 = 9

NCERT Solutions for Exercise 11.1 Class 11 Maths Chapter 11 - Conic Section

Edited By Vishal kumar | Updated on Nov 13, 2023 04:04 PM IST

NCERT Solutions for Class 11 Maths Chapter 11 - Conic Sections Exercise 11.1- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 11: Conic Sections Exercise 11.1- In the previous chapter of NCERT book Class 11 Mathematics, you have already learned about the lines, slope of lines, equation of lines in different forms, etc. In the NCERT solutions for Class 11 Maths chapter 11 exercise 11.1, you will learn about the circle and the equation of the circle in different forms. As you already learned about the circle and equations of circles in the previous classes, you will understand the NCERT syllabus Class 11 Maths chapter 11 exercise 11.1 very easily. Conic sections are obtained by intersections of a plane with a double-napped right circular cone.

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NCERT Solutions for Class 11 Maths Chapter 11 - Conic Sections Exercise 11.1- Download Free PDF
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Access Conic Section Class 11 Chapter 11 Exercise: 11.1
Question:1 Find the equation of the circle with
More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1:-
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NCERT Solutions for Class 11 Maths
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NCERT Solutions for Exercise 11.1 Class 11 Maths Chapter 11 - Conic Section

The first exercise of the chapter conic section is based on the section of a cone called the circle. You will get questions like finding the equation of a circle given the center and radius of the circle, finding the center and radius of the circle when the equation of the circle is given, etc in exercise 11.1 Class 11 Maths. You must try to solve all the NCERT problems by yourself before going through the Class 11 Maths chapter 11 exercise 11.1 solutions. If you are looking for NCERT solutions for Class 6 to Class 12, click on the NCERT Solutions link.

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Access Conic Section Class 11 Chapter 11 Exercise: 11.1

Question:1 Find the equation of the circle with

centre (0,2) and radius 2

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(0,2)$

AND $r=2$

So the equation of the circle is:

, $(x-0)^2+(y-2)^2=2^2$

$x^2+y^2-4y+4=4$

$x^2+y^2-4y=0$

Question:2 Find the equation of the circle with

centre (–2,3) and radius 4

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(-2,3)$

AND $r=4$

So the equation of the circle is:

, $(x-(-2))^2+(y-3)^2=4^2$

$x^2+4x+4+y^2-6y+9=16$

$x^2+y^2+4x-6y-3=0$

Question:3 Find the equation of the circle with

centre $\left(\frac{1}{2},\frac{1}{4} \right )$ and radius $\frac{1}{12}$

Answer:

As we know,

The equation of the circle with center ( h, k) and radius r is give by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=\left ( \frac{1}{2},\frac{1}{4} \right )$

AND

$r=\frac{1}{12}$

So the equation of circle is:

, $\left ( x-\frac{1}{2}\right )^2+\left ( y-\frac{1}{4}\right )^2=\left ( \frac{1}{12}\right )^2$

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

$x^2+y^2-x-\frac{1}{2}y-\frac{11}{36}=0$

$36x^2+36y^2-36x-18y-11=0$

Question:4 Find the equation of the circle with

centre (1,1) and radius $\sqrt2$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(1,1)$

AND $r=\sqrt{2}$

So the equation of the circle is:

, $(x-1)^2+(y-1)^2=(\sqrt{2})^2$

$x^2-2x+1+y^2-2y+1=2$

$x^2+y^2-2x-2y=0$

Question:5 Find the equation of the circle with

centre $(-a,-b)$ and radius $\sqrt{a^2 - b^2}$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So Given Here

$(h,k)=(-a,-b)$

AND $r=\sqrt{a^2-b^2}$

So the equation of the circle is:

, $(x-(-a))^2+(y-(-b))^2=(\sqrt{a^2-b^2})^2$

$x^2+2ax+a^2+y^2+2by+b^2=a^2-b^2$

$x^2+y^2+2ax+2by+2b^2=0$

Question:6 Find the centre and radius of the circles.

$(x+5)^2 + (y-3)^2 = 36$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$(x+5)^2 + (y-3)^2 = 36$

Can also be written in the form

$(x-(-5))^2 + (y-3)^2 = 6^2$

So, from comparing, we can see that

$r=6$

Hence the Radius of the circle is 6.

Question:7 Find the centre and radius of the circles.

$x^2 + y^2 -4x - 8y - 45 = 0$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$x^2 + y^2 -4x - 8y - 45 = 0$

Can also be written in the form

$(x-2)^2 + (y-4)^2 =(\sqrt{65})^2$

So, from comparing, we can see that

$r=\sqrt{65}$

Hence the Radius of the circle is $\sqrt{65}$ .

Question:8 Find the centre and radius of the circles.

$x^2 + y^2 -8x +10y -12 = 0$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$x^2 + y^2 -8x +10y -12 = 0$

Can also be written in the form

$(x-4)^2 + (y-(-5))^2 = (\sqrt{53})^2$

So, from comparing, we can see that

$r=\sqrt{53}$

Hence the radius of the circle is $\sqrt{53}$ .

Question:9 Find the centre and radius of the circles.

$2x^2 + 2y^2 - x = 0$

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given here

$2x^2 + 2y^2 - x = 0$

Can also be written in the form

$\left ( x-\frac{1}{4}\right )^2 + \left ( y-0 \right )^2 = \left ( \frac{1}{4} \right )^2$

So, from comparing, we can see that

$r=\frac{1}{4}$

Hence Center of the circle is the $\left ( \frac{1}{4},0\right )$ Radius of the circle is $\frac{1}{4}$ .

Question:10 Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line $4x + y = 16$ .

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given Here,

Condition 1: the circle passes through points (4,1) and (6,5)

$(4-h)^2+(1-k)^2=r^2$

$(6-h)^2+(5-k)^2=r^2$

Here,

$(4-h)^2+(1-k)^2=(6-h)^2+(5-k)^2$

$(4-h)^2-(6-h)^2+(1-k)^2-(5-k)^2=0$

$(-2)(10-2h)+(-4)(6-2k)=0$

$-20+4h-24+8k=0$

$4h+8k=44$

Now, Condition 2:centre is on the line $4x + y = 16$ .

$4h+k=16$

From condition 1 and condition 2

$h=3,\:k=4$

Now lets substitute this value of h and k in condition 1 to find out r

$(4-3)^2+(1-4)^2=r^2$

$1+9=r^2$

$r=\sqrt{10}$

So now, the Final Equation of the circle is

$(x-3)^2+(y-4)^2=(\sqrt{10})^2$

$x^2-6x+9+y^2-8y+16=10$

$x^2+y^2-6x-8y+15=0$

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

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

Given Here,

Condition 1: the circle passes through points (2,3) and (–1,1)

$(2-h)^2+(3-k)^2=r^2$

$(-1-h)^2+(1-k)^2=r^2$

Here,

$(2-h)^2+(3-k)^2=(-1-h)^2+(1-k)^2$

$(2-h)^2-(-1-h)^2+(3-k)^2-(1-k)^2=0$

$(3)(1-2h)+(2)(4-2k)=0$

$3-6h+8-4k=0$

$6h+4k=11$

Now, Condition 2: centre is on the line. $x - 3y - 11 = 0$

$h-3k=11$

From condition 1 and condition 2

$h=\frac{7}{2},\:k=\frac{-5}{2}$

Now let's substitute this value of h and k in condition 1 to find out $r$

$\left ( 2-\frac{7}{2}\right )^2+\left (3+\frac{5}{2}\right )^2=r^2$

$\frac{9}{4}+\frac{121}{4}=r^2$

$r^2=\frac{130}{4}$

So now, the Final Equation of the circle is

$\left(x-\frac{7}{2}\right )^2+\left(y+\frac{5}{2}\right)^2=\frac{130}{4}$

$x^2-7x+\frac{49}{4}+y^2+5y+\frac{25}{4}=\frac{130}{4}$

$x^2+y^2-7x+5y-\frac{56}{4}=0$

$x^2+y^2-7x+5y-14=0$

Question:12 Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

$(x-h)^2+(y-k)^2=r^2$

So let the circle be,

$(x-h)^2+(y-k)^2=r^2$

Since it's radius is 5 and its centre lies on x-axis,

$(x-h)^2+(y-0)^2=5^2$

And Since it passes through the point (2,3).

$(2-h)^2+(3-0)^2=5^2$

$(2-h)^2=25-9$

$(2-h)^2=16$

$(2-h)=4\:or\:(2-h)=-4$

$h=-2\: or\;6$

When $h=-2\:$ ,The equation of the circle is :

$(x-(-2))^2+(y-0)^2=5^2$

$x^2+4x+4+y^2=25$

$x^2+y^2+4x-21=0$

When $h=6$ The equation of the circle is :

$(x-6)^2+(y-0)^2=5^2$

$x^2-12x+36+y^2=25$

$x^2+y^2-12x+11=0$

Question:13 Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Answer:

Let the equation of circle be,

$(x-h)^2+(y-k)^2=r^2$

Now since this circle passes through (0,0)

$(0-h)^2+(0-k)^2=r^2$

$h^2+k^2=r^2$

Now, this circle makes an intercept of a and b on the coordinate axes.it means circle passes through the point (a,0) and (0,b)

So,

$(a-h)^2+(0-k)^2=r^2$

$a^2-2ah+h^2+k^2=r^2$

$a^2-2ah=0$

$a(a-2h)=0$

$a=0\:or\:a-2h=0$

Since $a\neq0\:so\:a-2h=0$

$h=\frac{a}{2}$

Similarly,

$(0-h)^2+(b-k)^2=r^2$

$h^2+b^2-2bk+k^2=r^2$

$b^2-2bk=0$

$b(b-2k)=0$

Since b is not equal to zero.

$k=\frac{b}{2}$

So Final equation of the Circle ;

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

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

$x^2+y^2-ax-bx=0$

Question:14 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Answer:

Let the equation of circle be :

$(x-h)^2+(y-k)^2=r^2$

Now, since the centre of the circle is (2,2), our equation becomes

$(x-2)^2+(y-2)^2=r^2$

Now, Since this passes through the point (4,5)

$(4-2)^2+(5-2)^2=r^2$

$4+9=r^2$

$r^2=13$

Hence The Final equation of the circle becomes

$(x-2)^2+(y-2)^2=13$

$x^2-4x+4+y^2-4y+4=13$

$x^2+y^2-4x-4y-5=0$

Question:15 Does the point (–2.5, 3.5) lie inside, outside or on the circle $x^2 + y^2 = 25$ ?

Answer:

Given, a circle

$x^2 + y^2 = 25$

As we can see center of the circle is ( 0,0)

Now the distance between (0,0) and (–2.5, 3.5) is

$d=\sqrt{(-2.5-0)^2+(3.5-0)^2}$

$d=\sqrt{6.25+12.25}$

$d=\sqrt{18.5}\approx 4.3$ $d=\sqrt{18.5}\approx 4.3<5$

Since distance between the given point and center of the circle is less than radius of the circle, the point lie inside the circle.

Key Features of Class 11 Maths Chapter 11 Exercise 11.1 Solution

Comprehensive Coverage: The ex 11.1 class 11 solution cover all exercise problems in Chapter 11.1 of the Class 11 Mathematics textbook.
Step-by-Step Solutions: Class 11 maths ex 11.1 solutions are presented in a clear, step-by-step format for easier understanding and application of problem-solving methods.
Clarity and Precision: Written with clarity and precision, ensuring easy comprehension of mathematical concepts and techniques.
Proper Mathematical Notation: Uses appropriate mathematical notations and terminology to enhance fluency in mathematical language.
Conceptual Understanding: 11th class maths exercise 11.1 answers aim to deepen conceptual understanding rather than promoting rote memorization, encouraging critical thinking.
Free Accessibility: Class 11 ex 11.1 solutions are typically available free of charge, providing an accessible resource for self-study.
Supplementary Learning Tool: Serves as a supplementary resource to complement classroom instruction and support exam preparation.
Homework and Practice: Students can use these class 11 maths chapter 11 exercise 11.1 solutions to cross-verify their work, practice problem-solving, and enhance overall mathematical skills.

Also see-

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. What are the conics ?

The curves obtained as intersections of a plane with a double-napped right circular cone are called conics.

2. What is circle ?

The curve obtained from the set of all points in a plane that are equidistant from a fixed point in the plane is called a circle.

3. What is the radius of the circle ?

The distance between the centre to a point on the circle is called the radius of the circle.

4. Write the equation of the circle with centre at (0,0) and radius 3 unit.

Equation of the circle centre at (0,0) and radius 3 => x^2 + y^2 = 3^2

x^2 + y^2 = 9

5. Write the equation of the circle with centre at (1,2) and radius 2 unit.

Equation of the circle centre at (1,2) and radius 2 => (x-1)^2 + (y-2)^2 = 2^2

x^2 -2x + 1 + y^2 - 4y + 4= 4

x^2 -2x + y^2 - 4y + 1= 0

6. Find the centre and radius of the circle x^2 + y^2 + 4x + 6y – 12 = 0

Equation of the circle => x^2 + y^2 + 4x + 6y – 12 = 0

(x+2)^2 + (y+3)^2 – 25 = 0

(x+2)^2 + (y+3)^2 = 5^2

Centre(-2,-3) and radius = 5.

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