If the line $lx+my=1$ touches the circle $x^2+y^2=a^2$, prove that the point $(l,m)$ lies on a circle with radius $\dfrac{1}{a}$.
My Attempt:
The center and radius of the circle $x^2+y^2=a^2$ are $(0,0)$ and $a$, respectively. Since $lx+my-1=0$ touches the circle,
$$a=\left|\dfrac {l\cdot0+m\cdot0-1}{\sqrt {l^2+m^2}}\right| \implies a^2=\dfrac {1}{l^2+m^2}.$$
Best Answer
That's perfectly right if you are allowed to use synthetic geometry results, namely that the tangent is orthogonal to the radius at the point of contact, so the distance of the line from the center of the circle equals the radius.
If the proof must be purely analytic, it's a bit more complicated.
I suggest to write points on the line $lx+my=1$ in the form $$ x=-mt, \qquad y=lt+1/m $$ (the case where $m=0$ can be treated as a special case). The intersections with the given circle can be computed from $$ m^2t^2+l^2t^2+2\frac{l}{m}t+\frac{1}{m^2}-a^2=0 $$ and the discriminant of the polynomial must be $0$, so $$ \frac{l^2}{m^2}-(m^2+l^2)\left(\frac{1}{m^2}-a^2\right)=0 $$ which reduces to $$ -1+a^2(m^2+l^2)=0 $$ which is the thesis.
For $m=0$ the line is $lx=1$ and the intersection with the circle are $$ \frac{1}{l^2}+y^2=a^2 $$ Since this equation must be satisfied by a single point, we get $l^2=\frac{1}{a^2}$.