Question

If the line $lx+my+n=0$ touches a fixed circle $x^2+y^2+2gx+2fy+c=0$ such that $4l^2+3m^2+n^2+2mn=0$, then $|g+f+c|$ is equal to

Answer 1

If the line $lx+my+n=0$ touches a fixed circle then the length of perpendicular from centre = Radius
$\left|\frac{-lg-mf+n}{\sqrt{l^2+m^2}}\right|=\sqrt{g^2+f^2-c}$

Squaring both sides,
$\Rightarrow (lg+mf-n{)}^2=(g^2+f^2-c\left)\right(l^2+m^2)$
$\Rightarrow l^2{g}^2+m^2{f}^2+n^2+2mlgf-2fmn-2nlg$$=g^2{l}^2+g^2{m}^2+f^2{l}^2+f^2{m}^2-c{l}^2-c{m}^2$
Now compare the given relation with $4l^2+3m^2+n^2+2mn=0$, we get $|g+f+c|=4$.