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Vaibhav Mathur Grade: 12
        

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

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

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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Badiuddin

7 years ago
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