Answer
The families of curves are orthogonal trajectories of each other.
We can see a sketch of the families of curves below.
Work Step by Step
Choose any curve from each family of curves.
If $ax+by = 0$, then $y = -\frac{ax}{b}$
Suppose $(x_0,y_0)$ is a point of intersection of the two curves. Then this point of intersection has the form $(x_0,-\frac{ax_0}{b})$
$x^2+y^2 = r^2$
We can find $y'$:
$2x+2y~y' = 0$
$y' = -\frac{x}{y}$
A tangent line at a point of intersection $(x_0,y_0)$ has the slope $-\frac{x_0}{y_0} = -\frac{x_0}{-\frac{ax_0}{b}} = \frac{b}{a}$
$ax+by = 0$
We can find $y'$:
$a+b~y' = 0$
$y' = -\frac{a}{b}$
A tangent line at a point of intersection $(x_0,y_0)$ has the slope $-\frac{a}{b}$
Since the slopes are negative reciprocals of each other, the two tangent lines are perpendicular. Therefore, the curves are orthogonal, and the families of curves are orthogonal trajectories of each other.
We can see a sketch of the families of curves below.
