Our ncert solutions for Exercise 11.1 Class 11 maths 11. Conic Sections - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 11.1 Class 11 maths 11. Conic Sections - ncert solutions - Toppers Study

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## Exercise 11.1 Class 11 maths 11. Conic Sections - ncert solutions - Toppers Study

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### Exercise 11.1 class 11 Mathematics Chapter 11. Conic Sections

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- Exercise 11.1 Class 11 Maths 11. Conic Sections - Ncert Solutions - Toppers Study
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## 11. Conic Sections

### | Exercise 11.1 |

## Exercise 11.1 Class 11 maths 11. Conic Sections - ncert solutions - Toppers Study

## Exercise 11.1 (Conic Sections)

**Q1. Find the equation of the circle with centre (0, 2) and radius 2.**

**Solution:**

The equation of a circle with centre (h, k) and radius r is given as

(x – h)^{2} + (y – k)^{2} = r^{2}

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x – 0)^{2} + (y – 2)^{2} = 2^{2}

⟹ x^{2} + y^{2} + 4 – 4 y = 4

⟹ x^{2} + y^{2 }– 4y = 0

**Q2. Find the equation of the circle with centre (–2, 3) and radius 4.**

**Solution: **

The equation of a circle with centre (h, k) and radius r is given as

(x – h)^{2} + (y – k)^{2} = r^{2}

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.

Therefore, 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

Q3.

**Q10. 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 the equation of the required circle be (x – h)^{2} + (y – k)^{2} = r^{2}.

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

(4 – h)^{2} + (1 – k)^{2} = r^{2} …………………. (1)

(6 – h)^{2} + (5 – k)^{2} = r^{2} …………………. (2)

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

∴ 4h + k = 16 …………………………………… (3)

From equations (1) and (2), we obtain

(4 – h)^{2} + (1 – k)^{2} = (6 – h)^{2} + (5 – k)^{2}

⇒ 16 – 8h + h^{2} + 1 – 2k + k^{2} = 36 – 12h + h^{2} + 25 – 10k + k^{2}

⇒ 16 – 8h + 1 – 2k = 36 – 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} = r^{2}

⇒ (1)^{2} + (– 3)^{2} = r^{2}

⇒ 1 + 9 = r^{2}

⇒ r^{2} = 10

⇒ 𝑟=√10

Thus, the equation of the required circle is

(x – 3)^{2} + (y – 4)^{2} = (√10)^{2}

x2 – 6x + 9 + y2 – 8y + 16 = 10

x2 + y2 – 6x – 8y + 15 = 0

The required equation for given circle is x2 + y2 – 6x – 8y + 15 = 0

##### Other Pages of this Chapter: 11. Conic Sections

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