Question-45785

Question Number 45785 by Tawa1 last updated on 16/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18

l e t c e n t r e o f c i r c l e i s (α, β) a n d r a d i u s r

s o e q n o f c i r c l e {(x - α)}^{2} + {(y - β)}^{2} = r^{2}

n o w a s p e r c o n d i t i o n

1) x a x i s i s [t h e t a n g e n t o f c i r c l e s o d i s t a n c e

f r o m c e n t e t o [x s x i s i s t h e r a d i u s .

d i s t a n c e f r o m c e n t r e (α, β) t o x a x i s i s β = r a d i u s

β = r

s o e q n i s {(x - α)}^{2} + {(y - β)}^{2} = β^{2}

2) c e n t r e l i e s o n 2 y = x

2 β = α

3) (14, 2) l i e z o n c i r c l e s o

{(14 - α)}^{2} + {(2 - β)}^{2} = β^{2}

{(14 - 2 β)}^{2} + {(2 - β)}^{2} = β^{2} g i v e n α = 2 β

196 - 56 β + 4 β^{2} + 4 - 4 β + β^{2} = β^{2}

4 β^{2} - 60 β + 200 = 0

β^{2} - 15 β + 50 = 0

(β - 5) (β - 10) = 0

β = 5 α = 10 e q n {(x - 10)}^{2} + {(y - 5)}^{2} = 5^{2}

w h e n β = 10 α = 20

e q n {(x - 20)}^{2} + {(y - 10)}^{2} = 10^{2}

Commented by Tawa1 last updated on 16/Oct/18

God bless you sir

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