if 4l^2-5m^2+6l+1=0 then the line lx+my+1=0 touches a fixed circle of centre (3,0) and radiusa)3^1/2 b)2^1/2 c)5^1/2 d)7^1/2

Dear Vaibhav

4l²-5m²+6l+1=0

 5( l² +m²) =9l² +6l +1

or 5( l² +m²) =(3l+1)²

 or v5v( l² +m²) =|(3l+1)|

now length of perpendicular from(3,0)  on line lx +my +1 =0

         r = |3l+1|/v( l² +m²)

            from above rresult

    r = v5

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