5. Four lines $l_{1}: x+3 y-15=0, l_{2}: k x$ $-y-6=0, l_{3}: x+5 y=0, l_{4}: y=0$ form a quadrilateral. To make this quadrilateral have a circumscribed circle, the value of $k$ is $\qquad$ .
5. $-\frac{4}{7}$.

Let the equation of the linear system passing through the four vertices of the quadrilateral be $(x+3 y-15)(x+5 y)+\lambda(k x-y-6) y=0$. From $\left\{\begin{array}{l}15-\lambda=1, \\ 8+k \lambda=0\end{array}\right.$, we get $k=-\frac{4}{7}$.