NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Conic Sections |

Exercise |
Ex 11.1 |

Number of Questions Solved |
15 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1

**In each of the following Exercises 1 to 5, find the equation of the circle with**

**Ex 11.1 Class 11 Maths Question 1.**

centre (0, 2) and radius 2

**Solution:**

Here h = 0,k = 2 and r = 2

The equation of circle is,

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

∴ (x – 0)^{2} + (y – 2)^{2} = (2)^{2}

⇒ x^{2} + y^{2} + 4 – 4y = 4

⇒ x^{2} + y^{2} – 4y = 0

**Ex 11.1 Class 11 Maths Question 2.**

centre (-2,3) and radius 4

**Solution:**

Here h=-2,k = 3 and r = 4

The equation of circle is,

(x – h)^{2} + (y – k)^{2} = r^{2}

∴(x + 2)^{2} + (y – 3)^{2} = (4)^{2}

⇒ x^{2} + 4 + 4x + y^{2} + 9 – 6y = 16

⇒ x^{2} + y^{2} + 4x – 6y – 3 = 0

**Ex 11.1 Class 11 Maths Question 3.**

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

**Solution:**

here h = \(\frac { 1 }{ 2 } \), k = \(\frac { 1 }{ 4 } \) and r = \(\frac { 1 }{ 12 } \)

The equation of circle is,

**Ex 11.1 Class 11 Maths Question 4.**

centre (1, 1) and radius \(\sqrt { 2 } \)

**Solution:**

Here h = l, k=l and r = \(\sqrt { 2 } \)

The equation of circle is,

(x – h)^{2} + (y – k)^{2} = r^{2}

∴ (x – 1)^{2} + (y – 1)^{2} = \(\left( \sqrt { 2 } \right) \)^{2}

⇒ x^{2} + 1 – 2x + y^{2} +1 – 2y = 2

⇒ x^{2} + y^{2} – 2x – 2y = 0

**Ex 11.1 Class 11 Maths Question 5.**

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

**Solution:**

Here h=-a, k = -b and r = \(\sqrt { { a }^{ 2 }-{ b }^{ 2 } } \)

The equation of circle is, (x – h)^{2} + (y – k)^{2} = r^{2}

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

⇒ x^{2} + a^{2} + 2ax + y^{2} + b^{2} + 2by = a^{2} -b^{2}

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

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

**Ex 11.1 Class 11 Maths Question 6.**

(x + 5)^{2} + (y – 3)^{2} = 36

**Solution:**

The given equation of circle is,

(x + 5)^{2} + (y – 3)^{2} = 36

⇒ (x + 5)^{2} + (y – 3)^{2} = (6)^{2}

Comparing it with (x – h)2^{2} + (y – k)^{2} = r^{2}, we get

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

Thus the co-ordinates of the centre are (-5, 3) and radius is 6.

**Ex 11.1 Class 11 Maths Question 7.**

x^{2} + y^{2} – 4x – 8y – 45 = 0

**Solution:**

The given equation of circle is

x^{2} + y^{2} – 4x – 8y – 45 = 0

∴ (x^{2} – 4x) + (y^{2} – 8y) = 45

⇒ [x^{2} – 4x + (2)^{2}] + [y^{2} – 8y + (4)^{2}] = 45 + (2)^{2} + (4)^{2}

⇒ (x – 2)^{2} + (y – 4)^{2} = 45 + 4 + 16

⇒ (x – 2)^{2} + (y – 4)^{2} = 65

⇒ (x – 2)^{2} + (y – 4)^{2}= \(\left( \sqrt { 65 } \right) ^{ 2 }\)

Comparing it with (x – h)^{2} + (y – k)^{2} = r^{2}, we

have h = 2,k = 4 and r = \(\sqrt { 65 } \).

Thus co-ordinates of the centre are (2, 4) and radius is \(\sqrt { 65 } \).

**Ex 11.1 Class 11 Maths Question 8.**

x^{2} + y^{2} – 8x + 10y – 12 = 0

**Solution:**

The given equation of circle is,

x^{2} + y^{2} – 8x + 10y -12 = 0

∴ (x^{2} – 8x) + (y^{2} + 10y) = 12

⇒ [x^{2} – 8x + (4)^{2}] + [y^{2} + 10y + (5)^{2}] = 12 + (4)^{2} + (5)^{2}

⇒ (x – 4)^{2} + (y + 5)^{2} = 12 + 16 + 25

⇒ (x – 4)^{2} + (y + 5)^{2} = 53

⇒ (x – 4)^{2} + (y + 5)^{2} = \(\left( \sqrt { 53 } \right) ^{ 2 }\)

Comparing it with (x – h)^{2} + (y – k)^{2} = r^{2}, we have h = 4, k = -5 and r = \(\sqrt { 53 } \)

Thus co-ordinates of the centre are (4, -5) and radius is \(\sqrt { 53 } \).

**Ex 11.1 Class 11 Maths Question 9.**

2x^{2} + 2y^{2} – x = 0

**Solution:**

The given equation of circle is,

2x^{2} + 2y^{2} – x = 0

**Ex 11.1 Class 11 Maths 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.

**Solution:**

The equation of the circle is,

(x – h)^{2} + (y – k)^{2} = r^{2} ….(i)

Since the circle passes through point (4, 1)

∴ (4 – h)^{2} + (1 – k)^{2} = r^{2}

⇒ 16 + h^{2} – 8h + 1 + k^{2} – 2k = r^{2}

⇒ h^{2}+ k^{2} – 8h – 2k + 17 = r^{2} …. (ii)

Also, the circle passes through point (6, 5)

∴ (6 – h^{2} + (5 – k)^{2} = r^{2}

⇒ 36 + h^{2} -12h + 25 + k^{2} – 10k = r^{2}

⇒ h^{2} + k^{2} – 12h – 10kk + 61 = r^{2} …. (iii)

From (ii) and (iii), we have h^{2} + k^{2} – 8h – 2k +17

= h^{2} + k^{2}– 12h – 10k + 61

⇒ 4h + 8k = 44 => h + 2k = ll ….(iv)

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

∴ 4h + k = 16 …(v)

Solving (iv) and (v), we get h = 3 and k = 4.

Putting value of h and k in (ii), we get

(3)^{2} + (4)^{2} – 8 x 3 – 2 x 4 + 17 = r^{2}

∴ r^{2} = 10

Thus required equation of circle is

(x – 3)^{2} + (y – 4)^{2} = 10

⇒ x^{2} + 9 – 6x + y^{2} +16 – 8y = 10

⇒ x^{2} + y^{2} – 6x – 8y +15 = 0.

**Ex 11.1 Class 11 Maths Question 11.**

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

**Solution:**

The equation of the circle is,

(x – h)^{2} + (y – k)^{2} = r^{2} ….(i)

Since the circle passes through point (2, 3)

∴ (2 – h)^{2} + (3 – k)^{2} = r^{2}

⇒ 4 + h^{2} – 4h + 9 + k^{2} – 6k = r^{2}

⇒ h^{2}+ k^{2} – 4h – 6k + 13 = r^{2} ….(ii)

Also, the circle passes through point (-1, 1)

∴ (-1 – h)^{2} + (1 – k)^{2} = r^{2}

⇒ 1 + h^{2} + 2h + 1 + k^{2} – 2k = r^{2}

⇒ h^{2} + k^{2} + 2h – 2k + 2 = r^{2} ….(iii)

From (ii) and (iii), we have

h^{2} + k^{2} – 4h – 6k + 13 = h^{2} + k^{2} + 2h – 2k + 2

⇒ -6h – 4k = -11 ⇒ 6h + 4k = 11 …(iv)

Since the centre (h, k) of the circle lies on the line x – 3y-11 = 0.

∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v)

Solving (iv) and (v), we get

h = \(\frac { 7 }{ 2 } \) and k = \(\frac { -5 }{ 2 } \)

Putting these values of h and k in (ii), we get

\(\left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { 4\times 7 }{ 2 } -6\times \frac { -5 }{ 2 } +13={ r }^{ 2 }\)

⇒ \(\frac { 49 }{ 4 } +\frac { 25 }{ 4 } -14+15+13 \) ⇒ \({ r }^{ 2 }=\frac { 65 }{ 2 } \)

Thus required equation of circle is

⇒ \(\left( x-\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 } \)

⇒ \({ x }^{ 2 }+\frac { 49 }{ 4 } -7x+{ y }^{ 2 }+\frac { 25 }{ 4 } +5y=\frac { 65 }{ 2 } \)

⇒ 4x^{2} + 49 – 28x + 4y^{2} + 25 + 20y = 130

⇒ 4x^{2} + 4y^{2} – 28x + 20y – 56 = 0

⇒ 4(x^{2} + y^{2} – 7x + 5y -14) = 0

⇒ x^{2} + y^{2} – 7x + 5y -14 = 0.

**Ex 11.1 Class 11 Maths Question 12.**

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

**Solution:**

Since the centre of the circle lies on x-axis, the co-ordinates of centre are (h, 0). Now the circle passes through the point (2, 3).

∴ Radius of circle

**Ex 11.1 Class 11 Maths Question 13.**

Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.

**Solution:**

Let the circle makes intercepts a with x-axis and b with y-axis.

∴ OA = a and OB = b

So the co-ordinates of A are (a, 0) and B are (0,b)

Now, the circle passes through three points 0(0, 0), A(a, 0) and B(0, b).

**Ex 11.1 Class 11 Maths Question 14.**

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

**Solution:**

The equation of circle is

(x – h)^{2} + (y – k)^{2} = r^{2} ….(i)

Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2)

∴ radius of circle

**Ex 11.1 Class 11 Maths Question 15.**

Does the point (-2.5, 3.5) lie inside, outside or on the circle x^{2} + y^{2} = 25?

**Solution:**

The equation of given circle is x^{2} + y^{2} = 25

⇒ (x – 0)^{2} + (y – 0)^{2} = (5)^{2}

Comparing it with (x – h)^{2} + (y – k)^{2} = r^{2}, we

get

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

Now, distance of the point (-2.5, 3.5) from the centre (0, 0)

\(\sqrt { \left( 0+2.5 \right) ^{ 2 }+\left( 0-3.5 \right) ^{ 2 } } =\sqrt { 6.25+12.25 } \)

\(\sqrt { 18.5 } \) = 4.3 < 5.

Thus the point (-2.5, 3.5) lies inside the circle.

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