If 4a^2 −5b^2 +6a+1=0 and the line ax+by+1=0 touches a fixed circle then the correct statement is : 1) centre of circle is at (3,0) 2) radius of circle is (√3). 3) circle passes through (1,0). 4) none of these.


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Question Number 32679 by rahul 19 last updated on 31/Mar/18

$I f 4 a^{2} - 5 b^{2} + 6 a + 1 = 0 a n d t h e l i n e$ $a x + b y + 1 = 0 t o u c h e s a f i x e d c i r c l e$ $t h e n t h e c o r r e c t s t a t e m e n t i s :$ $1) c e n t r e o f c i r c l e i s a t (3, 0)$ $2) r a d i u s o f c i r c l e i s \sqrt{3} .$ $3) c i r c l e p a s s e s t h r o u g h (1, 0) .$ $4) n o n e o f t h e s e .$
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