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

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


a)3^1/2 b)2^1/2 c)5^1/2 d)7^1/2

Grade:12

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

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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