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Question Number 84157 by redmiiuser last updated on 10/Mar/20

i f a c i r c l e h a v i n g a n

e q u a t i o n x^{2} + y^{2} - 6 x - 8 y = 0

i s i n t e r s e c t e d a t A a n d B

b y x + y = 1 . f i n d t h e

e q u a t i o n o f t h e c i r c l e

o n A B a s d i a m e t e r

Answered by TANMAY PANACEA last updated on 10/Mar/20

x^{2} + {(1 - x)}^{2} - 6 x - 8 (1 - x) = 0

x^{2} + 1 - 2 x + x^{2} - 6 x - 8 + 8 x = 0

2 x^{2} - 7 = 0 x = \pm \sqrt{\frac{7}{2}}

A (\sqrt{\frac{7}{2}}, 1 - \sqrt{\frac{7}{2}}) B (- \sqrt{\frac{7}{2}}, 1 + \sqrt{\frac{7}{2}})

c e n t r e o f r e q u i r e d c i r c l e m i d p o i n t o f A B

(0, 1) d i a m e t e r = d i s t a n c e b e t w e e n A B

2 r = {[{(2 \sqrt{\frac{7}{2}})}^{2} + {(2 \sqrt{\frac{7}{2}})}^{2}]}^{\frac{1}{2}}

2 r = {[(4 \times \frac{7}{2} + 4 \times \frac{7}{2})]}^{\frac{1}{2}} = 2 \sqrt{7}

r = \sqrt{7}

e q n c i r c l e

{(x - 0)}^{2} + {(y - 1)}^{2} = {(\sqrt{7})}^{2}

x^{2} + y^{2} - 2 y - 6 = 0

Answered by john santu last updated on 10/Mar/20

the equation of circle

(x - \sqrt{\frac{7}{2}}) (x + \sqrt{\frac{7}{2}}) + (y - (1 + \sqrt{\frac{7}{2}})) (y - (1 - \sqrt{\frac{7}{2}})) = 0

x^{2} - \frac{7}{2} + y^{2} - 2 y - \frac{5}{2} = 0

x^{2} + y^{2} - 2 y - 6 = 0

Answered by TANMAY PANACEA last updated on 10/Mar/20

a n o t h e r w a y

t h e e q n o f c i r c l e

(x^{2} + y^{2} - 6 x - 8 y) + λ (x + y - 1) = 0

x^{2} + y^{2} + x (- 6 + λ) + y (- 8 + λ) - λ = 0

c e n t r e o f r e q u i r e d c i r l e i s

c o m p a r i n g w i t h x^{2} + y^{2} + 2 g x + 2 f y + c = 0

2 g x = (- 6 + λ) x \to 2 g = - 6 + λ g = (\frac{λ - 6}{2})

2 f y = (- 8 + λ) y \to 2 f = - 8 + λ f = (\frac{λ - 8}{2})

c e n t r e o f r e q u i r e d c i r c l e (- g, - f)

(\frac{6 - λ}{2}, \frac{8 - λ}{2}) w h i c h l i e s o n x + y = 1

s o \frac{6 - λ}{2} + \frac{8 - λ}{2} = 1 \to 7 - λ = 1 s o λ = 6

r e q i r e d c i r c l e

x^{2} + y^{2} + x (- 6 + 6) + y (- 8 + 6) - 6 = 0

x^{2} + y^{2} - 2 y - 6 = 0