Given the following bundle of circles:
$$x^2(1 + k) + y^2(1 + k) + x(24 + 8k) + y(-4 -4k) + 4k + 132$$
I would have to find the circles tangent to the x-axis.
The way I would solve this is: plug $0$ into the $y$ variable (intersect the bundle with the x-axis), solve the resulting second degree equation and again solve for $k$ by forcing the discriminant to $0$.
I don't understand why, however, this method works; I would never solve a problem by using a method I don't understand.
Intersecting the bundle of circles with the x-axis gives us a second degree equation, which is a parabola (a bundle of parabolas, actually). Now, generally, forcing the discriminant of a second degree equation to zero really means limiting the solutions to when they are only one and not two; more clearly, it means we're looking for only one intersection point.
Geometrically, on a graph, what does it mean to force the discriminant of the above equation to zero?
What does the (bundle of) parabola we found before represent?
If you have
$$ x^2 (1 + k) + y^2 (1 + k) + x (24 + 8 k) - 4y (1 + k) + 4 k + 132=0 $$
The intersection with $y = \mu$ is obtained by solving for $x$
$$ x^2 (1 + k) + \mu^2 (1 + k) + x (24 + 8 k) - 4\mu (1 + k) + 4 k + 132=0 $$
or
$$ x = \frac{\pm\sqrt{-(k (\mu -6)+\mu +2) (k (\mu +2)+\mu -6)}-4 k-12}{k+1} $$
but at tangency
$$ (k (\mu -6)+\mu +2) (k (\mu +2)+\mu -6) = 0 $$
then
$$ k (\mu -6)+\mu +2=0 \ \ \mbox{or}\ \ k (\mu +2)+\mu -6 = 0 $$
but we need the tangency when $\mu = 0$ then
$$ -6k +2=0 \ \ \mbox{or}\ \ 2k -6 = 0 $$
giving
$$ k = \frac 13\ \ \mbox{and} \ \ k = 3 $$
Attached a plot showing the circles bundle (light blue) and in red the circle for $k = \frac 13$
NOTE
The circles's bundle can be written as
$$ \left(x+\frac{4 (k+3)}{k+1}\right)^2+(y-2)^2=\left(4\frac{k-1}{k+1}\right)^2 $$