Login

Class-12-science»Maths

Suppose f (x, y)=0 is the equation of the circle such that f(x, 1)=0 has equal roots(each equal to 2) and f (1, x)=0 also has equal roots (each equal to 0). Find the equation of the circle. Thanks !

$L e t t h e e q u a t i o n o f c i r c l e, f (x, y) = x^{2} + y^{2} + 2 g x + 2 f y + c ..... (1) G i v e n, f (x, y) = 0, S o, x^{2} + y^{2} + 2 g x + 2 f y + c = 0 ..... (2) A n d, f (x, 1) = 0, s o e q u a t i o n (1) b e c o m e s, x^{2} + {(1)}^{2} + 2 g x + 2 f \times 1 + c = 0 \Rightarrow x^{2} + 2 g x + 2 f + c + 1 = 0 a n d h a s e q u a l r o o t s (e a c h e q u a l t o 2), s o, x^{2} + 2 g x + 2 f + c + 1 = {(x - 2)}^{2} \Rightarrow x^{2} + 2 g x + 2 f + c + 1 = x^{2} - 4 x + 4 C o m p a r i n g t h e c o e f f i c i e n t s o f x b o t h s i d e, w e h a v e, 2 g = - 4 \Rightarrow g = - 2 A n d, 2 f + c + 1 = 4 \Rightarrow 2 f + c = 3 \Rightarrow 2 f = 3 - c A n d, A n d, f (1, x) = 0, s o e q u a t i o n (1) b e c o m e s, {(1)}^{2} + {(y)}^{2} + 2 g \times 1 + 2 f x + c = 0 \Rightarrow y^{2} + 2 g + 2 f x + c + 1 = 0 \Rightarrow y^{2} + 2 \times - 2 + (3 - c) x + c + 1 = 0 (p u t g = - 2 a n d 2 f = 3 - c) \Rightarrow y^{2} + (3 - c) x + c - 3 = 0 S i n c e e a c h r o o t e q u a l t o z e r o, t h e r e f o r e, {(3 - c)}^{2} = 4 (1) (c - 3) \Rightarrow 9 + c^{2} - 6 c = 4 c - 12 \Rightarrow c^{2} - 10 c + 21 = 0 \Rightarrow (c - 3) (c - 7) = 0 \Rightarrow c = 3 o r c = 7 W h e n c = 3, 2 f = 3 - 3 = 0 \Rightarrow f = 0 S o, (g, f, c) = (- 2, 0, 3) T h u s, e q u a t i o n o f c i r c l e i s, x^{2} + y^{2} + 2 (- 2) x + 2 (0) y + (3) = 0 \Rightarrow x^{2} + y^{2} - 4 x + 3 = 0 W h e n c = 7, 2 f = 3 - 7 = - 4 \Rightarrow f = - 2 S o, (g, f, c) = (- 2, - 2, 7) T h u s, e q u a t i o n o f c i r c l e i s, x^{2} + y^{2} + 2 (- 2) x + 2 (- 2) y + (7) = 0 \Rightarrow x^{2} + y^{2} - 4 x - 4 y + 7 = 0$

42

View Full Answer

Copyright © 2024 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?