Let $l_1$ be given by $a_1x+b_1y+c_1=0$ and $l_1$ be given by $a_2x+b_2y+c_2=0$.
Let $P$ be the point $(h,k)$ and $d_j$ be the distance of the point $P$ from line $l_j$. Then,
$$d_1=\frac{|a_1h+b_1k+c_1|}{\sqrt{a_1^2+b_1^2}} \quad \text{ and } \quad d_2=\frac{|a_2h+b_2k+c_2|}{\sqrt{a_2^2+b_2^2}}$$
We are given that the locus of the points for which $d_1^2+d_2^2=s$ (where $s$ is some constant) is a circle. Let $\lambda_1=a_1^2+b_1^2$ and $\lambda_2=a_2^2+b_2^2$. Observe that
$$d_1^2+d_2^2=\left(\frac{\lambda_2a_1^2+\lambda_1a_2^2}{\lambda_1 \lambda_2}\right)h^2+\left(\frac{\lambda_2b_1^2+\lambda_1b_2^2}{\lambda_1 \lambda_2}\right)k^2+\left(\frac{2a_1b_1}{\lambda_1}+\frac{2a_2b_2}{\lambda_2}\right)hk+\dotsb$$
For this to be a circle,
- coefficient of $hk$ must be $0$, and
- coefficients of $h$ and $k$ should be equal.
Thus we have
\begin{align*}
\lambda_2a_1^2+\lambda_1a_2^2 & = \lambda_2b_1^2+\lambda_1b_2^2\\
a_1b_1\lambda_2+a_2b_2\lambda_1 & = 0
\end{align*}
Upon solving this, we get
$$(a_1a_2)^2 = (b_1b_2)^2 \implies a_1a_2 = \pm b_1b_2.$$
Try to see why $a_1a_2=b_1b_2$ will not occur. Then the only thing left is $a_1a_2+b_1b_2=0$ which is the condition for perpendicularity of $l_1$ and $l_2$.