Step 1: Understand the inversion idea.
A point $P$ moves on the line $lx+my+n=0$. Another point $Q$ sits on the line joining the origin $O$ and $P$. We are told $OP \cdot OQ = k^2$. This is the rule of an inversion. So $Q$ is just $P$ scaled along the same direction from the origin.
Step 2: Write $Q$ as a scaled version of $P$.
Let $P=(\alpha,\beta)$ and $Q=(x,y)$. Since both lie on a line through the origin, they point the same way. So we can write \[ (x,y)=\lambda(\alpha,\beta). \] This means $x=\lambda\alpha$ and $y=\lambda\beta$.
Step 3: Turn the distance rule into a value of $\lambda$.
Because $Q=\lambda P$, the length $OQ=\lambda\,OP$. The rule $OP\cdot OQ=k^2$ becomes \[ \lambda\,(OP)^2=k^2. \] Now $(OP)^2=\alpha^2+\beta^2$ and also $x^2+y^2=\lambda^2(\alpha^2+\beta^2)$, so $\alpha^2+\beta^2=\dfrac{x^2+y^2}{\lambda^2}$. Putting this in gives \[ \lambda\cdot\frac{x^2+y^2}{\lambda^2}=k^2 \quad\Rightarrow\quad \lambda=\frac{x^2+y^2}{k^2}. \]
Step 4: Get $P$ back in terms of $Q$.
From $x=\lambda\alpha$ we get $\alpha=\dfrac{x}{\lambda}$, and likewise $\beta=\dfrac{y}{\lambda}$. We will plug these into the line that $P$ obeys.
Step 5: Use the fact that $P$ lies on the given line.
Since $l\alpha+m\beta+n=0$, \[ l\frac{x}{\lambda}+m\frac{y}{\lambda}+n=0. \] Multiply through by $\lambda$: \[ lx+my+n\lambda=0. \]
Step 6: Replace $\lambda$ and clean up.
Put $\lambda=\dfrac{x^2+y^2}{k^2}$: \[ lx+my+n\cdot\frac{x^2+y^2}{k^2}=0. \] Multiply by $k^2$ and rearrange to get the locus of $Q$. \[ \boxed{n(x^2+y^2)=k^2(lx+my)} \]