If $4l^2-5m^2+6l+1=0$ and the line $lx+my+1=0$ touches a fixed circle, then 

Let the circle be ${(x-h)}^2+{(y-k)}^2=r^2$

then  $r=\frac{|lh+mk+1|}{\sqrt{(l^2+m^2)}}$

$or{l}^2(h^2-r^2)+m^2(k^2-r^2)+2lmhk+2lh+2mk+1=0$

also  $4l^2-5m^2+6l+1=0$

Comparing the coefficients of similar terms

$h^2-r^2=4,k^2-r^2=-5,hk=0$

$2h=6,2k=0$ $\therefore k=0,h=3and9-r^2=4$ $\therefore r=\sqrt{5}$

Hence, centre $(3,0)$ and  radius $\sqrt{5}$ .