NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections, contains solutions for all exercise 11.1 questions. The reason for practising NCERT Solution is always to score more in the final exam. Download NCERT Class 11 maths solutions to access the solutions for all the exercises of all the chapters.

A conic section is a chapter that deals with different sections of cones. There are a wide variety of topics covered in its first exercise. Exercise 11.1 of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections is based on the following topics:

- Introduction
- Sections of a Cone
- Circle, ellipse, parabola and hyperbola
- Degenerated conic sections

- Circle

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

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### Solutions for Class 11 Maths Chapter 11 â€“ Exercise 11.1

**In each of the following Exercise 1 to 5, find the equation of the circle with
1. Centre (0, 2) and radius 2**

**Solution:**

Given:

Centre (0, 2) and radius 2

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (0, 2) and radius (r) = 2

The equation of the circle is

(x â€“ 0)^{2 }+ (y â€“ 2)^{2 }= 2^{2}

x^{2 }+ y^{2 }+ 4 â€“ 4y = 4

x^{2 }+ y^{2}Â â€“ 4y = 0

âˆ´ The equation of the circle is x^{2 }+ y^{2}Â â€“ 4y = 0

**2. Centre (â€“2, 3) and radius 4**

**Solution:**

Given:

Centre (-2, 3) and radius 4

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (-2, 3) and radius (r) = 4

The equation of the circle is

(x + 2)^{2}Â + (y â€“ 3)^{2}Â = (4)^{2}

x^{2}Â + 4x + 4 + y^{2}Â â€“ 6y + 9 = 16

x^{2}Â + y^{2}Â + 4x â€“ 6y â€“ 3 = 0

âˆ´ The equation of the circle is x^{2}Â + y^{2}Â + 4x â€“ 6y â€“ 3 = 0

**3. Centre (1/2, 1/4) and radius (1/12)**

**Solution:**

Given:

Centre (1/2, 1/4) and radius 1/12

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (1/2, 1/4) and radius (r) = 1/12

The equation of the circle is

(x â€“ 1/2)^{2} + (y â€“ 1/4)^{2} = (1/12)^{2}

x^{2} â€“ x + Â¼ + y^{2} â€“ y/2 + 1/16 = 1/144

x^{2} â€“ x + Â¼ + y^{2} â€“ y/2 + 1/16 = 1/144

144x^{2}Â â€“ 144x + 36 + 144y^{2}Â â€“ 72y + 9 â€“ 1 = 0

144x^{2}Â â€“ 144x + 144y^{2}Â â€“ 72y + 44 = 0

36x^{2}Â + 36x + 36y^{2} â€“ 18y + 11 = 0

36x^{2}Â + 36y^{2}Â â€“ 36x â€“ 18y + 11= 0

âˆ´ The equation of the circle is 36x^{2}Â + 36y^{2}Â â€“ 36x â€“ 18y + 11= 0

**4. Centre (1, 1) and radius âˆš2**

**Solution:**

Given:

Centre (1, 1) and radius âˆš2

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (1, 1) and radius (r) = âˆš2

The equation of the circle is

(x-1)^{2}Â + (y-1)^{2 }= (âˆš2)^{2}

x^{2}Â â€“ 2x + 1 + y^{2}Â -2y + 1 = 2

x^{2}Â + y^{2}Â â€“ 2x -2y = 0

âˆ´ The equation of the circle is x^{2}Â + y^{2}Â â€“ 2x -2y = 0

**5. Centre (â€“a, â€“b) and radius âˆš(a ^{2} â€“ b^{2})**

**Solution:**

Given:

Centre (-a, -b) and radius âˆš(a^{2} â€“ b^{2})

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (-a, -b) and radius (r) = âˆš(a^{2} â€“ b^{2})

The equation of the circle is

(x + a)^{2}Â + (y + b)^{2}Â = (âˆš(a^{2} â€“ b^{2})^{2})

x^{2}Â + 2ax + a^{2}Â + y^{2}Â + 2by + b^{2}Â = a^{2}Â â€“ b^{2}

x^{2}Â + y^{2}Â +2ax + 2by + 2b^{2}Â = 0

âˆ´ The equation of the circle is x^{2}Â + y^{2}Â +2ax + 2by + 2b^{2}Â = 0

**In each of the following Exercise 6 to 9, find the centre and radius of the circles.**

6. (x + 5)^{2}Â + (y â€“ 3)^{2}Â = 36

**Solution:**

Given:

The equation of the given circle is (x + 5)^{2}Â + (y â€“ 3)^{2}Â = 36

(x â€“ (-5))^{2} + (y â€“ 3)^{2}Â = 6^{2} [which is of the form (x â€“ h)^{2}Â + (y â€“ k )^{2}Â = r^{2}]

Where, h = -5, k = 3 and r = 6

âˆ´ The centre of the given circle is (-5, 3) and its radius is 6.

**7. x ^{2}Â + y^{2}Â â€“ 4x â€“ 8y â€“ 45 = 0**

**Solution:**

Given:

The equation of the given circle is x^{2}Â + y^{2}Â â€“ 4x â€“ 8y â€“ 45 = 0.

x^{2}Â + y^{2}Â â€“ 4x â€“ 8y â€“ 45 = 0

(x^{2}Â â€“ 4x) + (y^{2}Â -8y) = 45

(x^{2} â€“ 2(x) (2) + 2^{2}) + (y^{2} â€“ 2(y) (4) + 4^{2}) â€“ 4 â€“ 16 = 45

(x â€“ 2)^{2}Â + (y â€“ 4)^{2}Â = 65

(x â€“ 2)^{2}Â + (y â€“ 4)^{2}Â = (âˆš65)^{2} [which is form (x-h)^{2}Â +(y-k)^{2}Â = r^{2}]

Where h = 2, K = 4 and r = âˆš65

âˆ´ The centre of the given circle is (2, 4) and its radius is âˆš65.

**8. x ^{2}Â + y^{2}Â â€“ 8x + 10y â€“ 12 = 0**

**Solution:**

Given:

The equation of the given circle is x^{2}Â + y^{2}Â -8x + 10y -12 = 0.

x^{2}Â + y^{2}Â â€“ 8x + 10y â€“ 12 = 0

(x^{2}Â â€“ 8x) + (y^{2 }+ 10y) = 12

(x^{2} â€“ 2(x) (4) + 4^{2}) + (y^{2} â€“ 2(y) (5) + 5^{2}) â€“ 16 â€“ 25 = 12

(x â€“ 4)^{2}Â + (y + 5)^{2}Â = 53

(x â€“ 4)^{2}Â + (y â€“ (-5))^{2}Â = (âˆš53)^{2} [which is form (x-h)^{2}Â +(y-k)^{2}Â = r^{2}]

Where h = 4, K= -5 and r = âˆš53

âˆ´ The centre of the given circle is (4, -5) and its radius is âˆš53.

**9. 2x ^{2} + 2y^{2} â€“ x = 0**

**Solution:**

The equation of the given of the circle is 2x^{2}Â + 2y^{2}Â â€“x = 0.

2x^{2}Â + 2y^{2}Â â€“x = 0

(2x^{2}Â + x) + 2y^{2}Â = 0

(x^{2} â€“ 2 (x) (1/4) + (1/4)^{2}) + y^{2} â€“ (1/4)^{2} = 0

(x â€“ 1/4)^{2} + (y â€“ 0)^{2} = (1/4)^{2} [which is form (x-h)^{2}Â +(y-k)^{2}Â = r^{2}]

Where, h = Â¼, K = 0, and r = Â¼

âˆ´ The center of the given circle is (1/4, 0)Â and its radius isÂ 1/4.

**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.**

**Solution:**

Let us consider the equation of the required circle be (x â€“ h)^{2}+ (y â€“ k)^{2}Â = r^{2}

We know that the circle passes through points (4,1) and (6,5)

So,

(4 â€“ h)^{2 }+ (1 â€“ k)^{2}Â = r^{2}Â â€¦â€¦â€¦â€¦â€¦..(1)

(6â€“ h)^{2}+ (5 â€“ k)^{2}Â = r^{2}Â â€¦â€¦â€¦â€¦â€¦â€¦(2)

Since, the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k =16â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

From the equation (1) and (2), we obtain

(4 â€“ h)^{2}+ (1 â€“ k)^{2}Â =(6 â€“ h)^{2}Â + (5 â€“ k)^{2}

16 â€“ 8h + h^{2}Â +1 -2k +k^{2}Â = 36 -12h +h^{2}+15 â€“ 10k + k^{2}

16 â€“ 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k =11â€¦â€¦â€¦â€¦â€¦. (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 â€“ 3)^{2}+ (1 â€“ 4)^{2}Â = r^{2}

(1)^{2}Â + (-3)^{2}Â = r^{2}

1+9 = r^{2}

r =Â âˆš10

so now, (x â€“ 3)^{2 }+ (y â€“ 4)^{2}Â = (âˆš10)^{2}

x^{2}Â â€“ 6x + 9 + y^{2}Â â€“ 8y + 16 =10

x^{2}Â + y^{2}Â â€“ 6x â€“ 8y + 15 = 0

âˆ´ The equation of the required circle is x^{2}Â + y^{2}Â â€“ 6x â€“ 8y + 15 = 0

**11. Find the equation of the circle passing through the points (2, 3) and (â€“1, 1) and whose centre is on the line x â€“ 3y â€“ 11 = 0.**

**Solution: **

Let us consider the equation of the required circle be (x â€“ h)^{2 }+ (y â€“ k)^{2}Â = r^{2}

We know that the circle passes through points (2,3) and (-1,1).

(2 â€“ h)^{2}+ (3 â€“ k)^{2}Â =r^{2}Â â€¦â€¦â€¦â€¦â€¦..(1)

(-1 â€“ h)^{2}+ (1â€“ k)^{2}Â =r^{2}Â â€¦â€¦â€¦â€¦â€¦â€¦(2)

Since, the centre (h, k) of the circle lies on line x â€“ 3y â€“ 11= 0,

h â€“ 3k =11â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

From the equation (1) and (2), we obtain

(2 â€“ h)^{2}+ (3 â€“ k)^{2}Â =(-1 â€“ h)^{2}Â + (1 â€“ k)^{2}

4 â€“ 4h + h^{2}Â +9 -6k +k^{2}Â = 1 + 2h +h^{2}+1 â€“ 2k + k^{2}

4 â€“ 4h +9 -6k = 1 + 2h + 1 -2k

6h + 4k =11â€¦â€¦â€¦â€¦â€¦. (4)

Now let us multiply equation (3) by 6 and subtract it from equation (4) to get,

6h+ 4k â€“ 6(h-3k) = 11 â€“ 66

6h + 4k â€“ 6h + 18k = 11 â€“ 66

22 k = â€“ 55

K = -5/2

Substitute this value of K in equation (4) to get,

6h + 4(-5/2) = 11

6h â€“ 10 = 11

6h = 21

h = 21/6

h = 7/2

We obtain h =Â 7/2and k = -5/2

On substituting the values of h and k in equation (1), we get

(2 â€“ 7/2)^{2} + (3 + 5/2)^{2} = r^{2}

^{2}+ [(6+5)/2]

^{2}= r

^{2}

(-3/2)^{2} + (11/2)^{2} = r^{2}

9/4 + 121/4 = r^{2}

130/4 = r^{2}

The equation of the required circle is

(x â€“ 7/2)^{2} + (y + 5/2)^{2} = 130/4

^{2}+ [(2y+5)/2]

^{2}= 130/4

4x^{2}Â -28x + 49 +4y^{2}Â + 20y + 25 =130

4x^{2}Â +4y^{2}Â -28x + 20y â€“ 56 = 0

4(x^{2}Â +y^{2}Â -7x + 5y â€“ 14) = 0

x^{2 }+ y^{2 }â€“ 7x + 5y â€“ 14 = 0

âˆ´ The equation of the required circle is x^{2 }+ y^{2 }â€“ 7x + 5y â€“ 14 = 0

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

**Solution:**

Let us consider the equation of the required circle be (x â€“ h)^{2}+ (y â€“ k)^{2}Â = r^{2}

We know that the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

So now, the equation of the circle is (x â€“ h)^{2}Â + y^{2}Â = 25.

It is given that the circle passes through the point (2, 3) so the point will satisfy the equation of the circle.

(2 â€“ h)^{2}+ 3^{2}Â = 25

(2 â€“ h)^{2}Â = 25-9

(2 â€“ h)^{2}Â = 16

2 â€“ h = **Â±** **âˆš**16 = **Â±** 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}Â + y^{2}Â = 25

x^{2}Â + 12x + 36 + y^{2}Â = 25

x^{2}Â + y^{2}Â + 4x â€“ 21 = 0

When h = 6, the equation of the circle becomes

(x â€“ 6)^{2}Â + y^{2}Â = 25

x^{2}Â -12x + 36 + y^{2}Â = 25

x^{2}Â + y^{2}Â -12x + 11 = 0

âˆ´ The equation of the required circle is x^{2}Â + y^{2}Â + 4x â€“ 21 = 0 and x^{2}Â + y^{2}Â -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 â€“ h)^{2}+ (y â€“ k)^{2}Â =r^{2}

We know that the circle passes through (0, 0),

So, (0 â€“ h)^{2}+ (0 â€“ k)^{2}Â = r^{2}

h^{2}Â + k^{2}Â =Â r^{2}

Now, The equation of the circle is (x â€“ h)^{2 }+ (y â€“ k)^{2}Â = h^{2}Â + k^{2}.

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 â€“ h)^{2}+ (0 â€“ k)^{2}Â =h^{2}Â +k^{2}â€¦â€¦â€¦â€¦â€¦..(1)

(0 â€“ h)^{2}+ (bâ€“ k)^{2}Â =h^{2}Â +k^{2}â€¦â€¦â€¦â€¦â€¦â€¦(2)

From equation (1), we obtain

a^{2}Â â€“ 2ah + h^{2}Â +k^{2}Â = h^{2}Â +k^{2}

a^{2}Â â€“ 2ah = 0

a(a â€“ 2h) =0

a = 0 or (a -2h) = 0

However, a â‰ Â 0; hence, (a -2h) = 0

h = a/2

From equation (2), we obtain

h^{2}Â â€“ 2bk + k^{2}Â + b^{2}= h^{2}Â +k^{2}

b^{2}Â â€“ 2bk = 0

b(bâ€“ 2k) = 0

b= 0 or (b-2k) =0

However, aÂ â‰ 0; hence, (b -2k) = 0

kÂ =b/2

So, the equation is

(x â€“ a/2)^{2} + (y â€“ b/2)^{2} = (a/2)^{2} + (b/2)^{2}

^{2}+ [(2y-b)/2]

^{2}= (a

^{2}+ b

^{2})/4

4x^{2}Â â€“ 4ax + a^{2}Â +4y^{2}Â â€“ 4by + b^{2}Â = a^{2}Â + b^{2}

4x^{2}Â + 4y^{2}Â -4ax â€“ 4by = 0

4(x^{2}Â +y^{2}Â -7x + 5y â€“ 14) = 0

x^{2 }+ y^{2 }â€“ axÂ â€“ by = 0

âˆ´ The equation of the required circle is x^{2 }+ y^{2 }â€“ axÂ â€“ by = 0

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

**Solution:**

Given:

The centre of the circle is given as (h, k) = (2,2)

We know that the circle passes through point (4,5), the radius (r) of the circle is the distance between the points (2,2) and (4,5).

r = **âˆš**[(2-4)^{2} + (2-5)^{2}]

= âˆš[(-2)^{2} + (-3)^{2}]

= âˆš[4+9]

= âˆš13

The equation of the circle is given as

(xâ€“ h)^{2}+ (y â€“ k)^{2}Â = r^{2}

(x â€“h)^{2}Â + (y â€“ k)^{2}Â = (âˆš13)^{2}

(x â€“2)^{2}Â + (y â€“ 2)^{2}Â = (âˆš13)^{2}

x^{2}Â â€“ 4x + 4 + y^{2 }â€“ 4y + 4 = 13

x^{2}Â + y^{2}Â â€“ 4x â€“ 4y = 5

âˆ´ The equation of the required circle is x^{2}Â + y^{2}Â â€“ 4x â€“ 4y = 5

**15. Does the point (â€“2.5, 3.5) lie inside, outside or on the circle x ^{2}Â + y^{2}Â = 25?**

**Solution:**

Given:

The equation of the given circle isÂ x^{2}Â +y^{2}Â = 25.

x^{2}Â + y^{2}Â = 25

(x â€“ 0)^{2}Â + (y â€“ 0)^{2}Â = 5^{2} [which is of the form (x â€“ h)^{2 }+ (y â€“ k)^{2}Â = r^{2}]

Where, h = 0, k = 0 and r = 5.

So the distance between point (-2.5, 3.5) and the centre (0,0) is

= âˆš[(-2.5 â€“ 0)^{2} + (-3.5 â€“ 0)^{2}]

= âˆš(6.25 + 12.25)

= âˆš18.5

= 4.3 [which is < 5]

Since, the distance between point (-2.5, -3.5) and the centre (0, 0) of the circle is less than the radius of the circle, point (-2.5, -3.5) lies inside the circle.

### Access Other Exercise Solutions of Class 11 Maths Chapter 11- Conic Sections

Exercise 11.2 Solutions 12 Questions

Exercise 11.3 Solutions 20 Questions

Exercise 11.4 Solutions 15 Questions

Miscellaneous Exercise On Chapter 11 Solutions 8 Questions