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.

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

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