Answer : True
The equation of the given circle is $x^2+y^2=a^2$.
Hence its centre lies on the origin $O(0,0)$
Hence the perpendicular distance of the given line $lx+my-1=0$ is the radius.
$\therefore r=\bigg|\large\frac{l(0)+m(0)-1}{\sqrt{l^2+m^2}}\bigg|=\frac{1}{\sqrt{l^2+m^2}}$-----(1)
Condition for a line $y=mx+c$ to be a tangent to a circle is $r=\large\frac{c}{\sqrt{m^2+1}}$
Where $c$ is the intercept and $m$ is the slope of the line.
Here in the line $lx+my-1$;slope $m=-\large\frac{l}{m}$
$c=\large\frac{1}{m}$
$\therefore r=\large\frac{\Large\frac{1}{m}}{\sqrt{(-\Large\frac{l}{m})^2+1}}=\frac{1}{\sqrt{l^2+m^2}}$------(2)
Equ(1) = Equ(2)
Hence the statement is True.