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Question

Consider the relation 4l25m2+6l+1=0, where l,mR. If the line lx+my+1=0 touches a fixed circle, then the centre of that circle is: 
  1. (0,5)
  2. (3,0)
  3. (2,1)
  4. (3,3)


Solution

The correct option is B (3,0)
Let equation of circle be
x2+y2+2gx+2fy+c=0   (1)

The line lx+my+1=0, will touch circle if,
Length of perpendicular from centre = Radius ,
i.e, glmf+1l2+m2=g2+f2c 
(gl+mf1)2=(l2+m2)(g2+f2c)
(cf2)l2+(cg2)m22gl2fm+2gflm+1=0(2)

But the given relation on l,m is
4l25m2+6l+1=0   (3)
Comparing  equations (2),(3), we get
cf2=4,cg2=5,2g=6,2f=0,2gf=0
On solving we get,
f=0, g=3, c=4

centre (3,0)

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