Question

Consider the relation $4l^2-5m^2+6l+1=0$, where $l,m\in R$, then the line $lx+my+1=0$ touches a fixed circle whose centre and radius of circle are

• A. $(2,0),3$
• B. $(-3,0),\sqrt{3}$
• C. $(3,0),\sqrt{5}$
• D. $(-2,0),3$

Answer 1

The correct option is C $(3,0),\sqrt{5}$

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0\left(i\right)$
The line $lx+my+1=0$ will touch circle $\left(i\right)$ if the length of perpendicular from the centre $(-g,-f)$ of the circle on the line is equal to its radius, i.e.
$\frac{|-gl-mf+1|}{\sqrt{l^2+m^2}}=\sqrt{g^2+f^2-c}$
$(gl+mf-1{)}^2=(l^2+m^2\left)\right(g^2+f^2-c)$
or $(c-f^2)l^2+(c-g^2)m^2-2gl-2fm+2gflm+1=0\left(ii\right)$
But the given condition of tangency is $4l^2-5m^2+6l+1=0\left(iii\right)$
Comparing $\left(ii\right)$ and $\left(iii\right)$, we get
$c-f^2=4,c-g^2=-5,-2g=6,-2f=0,2gf=0$
Solving, we get
$f=0,g=-3,c=4$
Substituting these values in $\left(i\right)$, the equation of the circle is $x^2+y^2-6x+4=0$.
$\Rightarrow (x-3{)}^2+y^2=5$
Centre=$(3,0)$, radius $=\sqrt{5}$