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ANIMESH SAXENA Grade: 12
        

a line "lx+my+1=0", touches a fixed cirlce and satisfies the relation, "4l^2-5m^2+6l+1=0". find the centre of the cirlce.

6 years ago

Answers : (1)

Askiitians_Expert Yagyadutt
askIITians Faculty
74 Points
										

Hii Animesh ...!


 


Well this kind of question needs a lot practice on family of curves...otherwise..you will get confused when it comes in exam...So better refer any good book of co-ordinate geometry and solve some 20-30 question on family of curve....Okk


Now attempting your question !


 


Simply it needs a simple modification to find the equation of a circle using ...lx + my + 1 = 0  and the relation ..


4l^2-5m^2+6l+1=0


Now ...4l^2-5m^2+6l+1= ( lx + my + 1 )^2


 


And finally simplify it ..you will get a equation of circle..somewhat like this ....


 


l^2.x^2 + m^2.y^2 + 2.l.x + 2.m.y + 2.l.m.xy - 4.l^2 + 5.m^2 -6.l = 0


 


So definitely ...Centre will be ( -l,-m)  


 


I hope it is clear to you now...Tell me is the answer is right or not !


 


regards


Yagya


askiitians_expert



6 years ago
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