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CLASS_11_MATHS_SOLUTIONS_NCERT

Class XI Chapter

Class XI Chapter 11 – Conic Section Maths ______________________________________________________________________________ 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 = r2 ⇒ 1 + 9 = r 2 ⇒ r 2 = 10 ⇒ r = 10 Thus, the equation of the required circle is (x – 3) 2 + (y – 4) 2 = 2 10 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 whose centre is on the line x- 3y - 11 = 0. Solution 11: Let the equation of the required circle be (x – h) 2 + (y – k) 2 = r 2 . Since the circle passes through points (2, 3) and (–1, 1), (2 – h) 2 + (3 – k) 2 = r 2 … (1) (–1 – h) 2 + (1 – k) 2 = r 2 … (2) Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0, h – 3k = 11 … (3) From equations (1) and (2), we obtain (2 – h) 2 + (3 – k) 2 = (–1 – h) 2 + (1 – k) 2 ⇒ 4 – 4h + h 2 + 9 – 6k + k 2 = 1 + 2h + h 2 + 1 – 2k + k 2 ⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k ⇒ 6h + 4k = 11 … (4) Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

Class XI Chapter 11 – Conic Section Maths ______________________________________________________________________________ On solving equations (3) and (4), we obtain h = 7 2 and k = 5 2 On substituting the values of h and k in equation (1), we obtain 2 2 7 5 2 3 r 2 2 2 2 47 65 r 2 2 2 2 3 11 2 2 r 2 9 121 2 r 4 4 130 2 r 4 Thus, the equation of the required circle is 2 2 7 5 130 x y 2 2 4 2 2 2x7 2y5 130 2 2 4 2 2 4x 28x 49 4y 20y 25 130 2 2 4x 4y 28x 20y 56 0 2 2 4 x y 7x 5y 14 0 2 2 x y x y 7 5 14 0 2 2 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 12: Let the equation of the required circle be (x – h) 2 + (y – k) 2 = r 2 . Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5. Now, the equation of the circle becomes (x – h) 2 + y 2 = 25. It is given that the circle passes through point (2, 3). Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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    Class XI Chapter 2 - Relations and

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    Class XI Chapter 3 - Trigonometric

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    Class XI Chapter 4 - Principle of M

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    Class XI Chapter 5 - Complex Number

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    Class XI Chapter 6 - Linear Inequal

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    Class XI Chapter 7 - Permutation an

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    Class XI Chapter 8 - Binomial Theor

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    Class XI Chapter 9 - Sequences and

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    Class XI Chapter 10 - Straight Line

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    Class XI Chapter 12 - Introduction

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    Class XI Chapter 13 - Limits and De

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    Class XI Chapter 14 - Mathematical

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    Class XI Chapter 15 - Statistics Ma

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