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

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 11 |

Subject | Maths |

Chapter | Chapter 11 |

Exercise | Class 11 Maths Chapter 11 Conic Sections Exercise 11.1 |

Number of Questions Solved |

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

**NCERT TEXTBOOK EXERCISES**

**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 1/12

**Solution:**

here h = 1/2, k = 1/4 and r = 1/12

The equation of circle is,

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

centre (1, 1) and radius √2

**Solution:**

Here h = l, k=l and r = √2

The equation of circle is,

(x – h)^{2} + (y – k)^{2} = r^{2}

∴ (x – 1)^{2} + (y – 1)^{2} = (√2)^{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}= (√65)^{2}

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

have h = 2,k = 4 and r = √65.

Thus co-ordinates of the centre are (2, 4) and radius is √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} = (√53)^{2}

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

Thus co-ordinates of the centre are (4, -5) and radius is √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 = 7/2 and k = −5/2

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

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

⇒ $\frac { 49 }{ 4 } +\frac { 25 }{ 4 } -14+15+13$

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)

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

√18.5 = 4.3 < 5.

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

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

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