Question

Consider the relation $4l^2-5m^2+6l+1=0$, where $l,m\in R.$ If the line $lx+my+1=0$ touches a fixed circle, then the centre of that circle is:

• A. $(0,5)$
• B. $(3,0)$
• C. $(2,1)$
• D. $(3,-3)$

Answer 1

The correct option is B $(3,0)$

Let equation of circle be
$x^2+y^2+2gx+2fy+c=0\cdots \left(1\right)$

The line $lx+my+1=0$, will touch circle if,
Length of perpendicular from centre = Radius ,
i.e, $\left|\frac{-gl-mf+1}{\sqrt{l^2+m^2}}\right|=\sqrt{g^2+f^2-c}$
$\Rightarrow (gl+mf-1{)}^2=(l^2+m^2\left)\right(g^2+f^2-c)$
$\Rightarrow (c-f^2)l^2+(c-g^2)m^2-2gl-2fm+2gflm+1=0\cdots \left(2\right)$

But the given relation on $l,m$ is
$4l^2-5m^2+6l+1=0\cdots \left(3\right)$
Comparing equations $\left(2\right),\left(3\right)$, we get
$c-f^2=4,c-g^2=-5,-2g=6,-2f=0,2gf=0$
On solving we get,
$f=0,g=-3,c=4$

$\therefore$ centre $(3,0)$