From an example in the book,
Let $\rm\lambda C + \mu D$ be a family of circles with radical axis on as $x = 0$ and centre line as $y = 0$, Then the family can be described by $$y^2 + (x - \nu)^2 = \nu^2 - c$$ Where $\nu$ depends on $\lambda$ and $\mu$ and $c$ is a constant.
If $c > 0$ (say $ c = k^2$) then the circles are real only if $-k < \nu < k$ and all of them fail to intersect radical axis at any point.
Lets say $\nu = k - \varepsilon$ where $\varepsilon$ is a sufficiently small positive real number for which $-k < \nu < k$ is true, then
$$y^2 + (x - \nu)^2 = k^2 + \varepsilon^2 - 2k\varepsilon - k^2 = (\varepsilon - 2k)\varepsilon \implies \varepsilon \gt 2k \implies \nu < -k $$,which contracdicts our assumption.
So I think the right statement should be $|\nu| > k$ not $|\nu| < k$ as in the book.
Also if, for some choice of $c$ and $\nu$, $\nu^2 - c = \eta > 0$ then $$y^2 + (x - \nu)^2 = \eta$$
Then this circle will intersect $ y = 0 $ at $\left(\sqrt{\eta} + \nu, 0\right)$. Therefore all real circles in this family with $c > 0$ will intersect $y = 0$ (radical axis) at $\left(\sqrt{\nu^2 - c} + \nu, 0\right)$. Contradictory to what is written in the book.
Am I missing something or there is really two big typos in the example above ?
The family of circles has equation $x^2+y^2-2 \nu x+k^2=0$
Circles have centre at $(\nu,\;0)$ and radii $r=\sqrt{\nu^2-k^2}$ which are real iff
$\nu^2-k^2\ge 0$ that is $\nu\le -k\lor \nu \ge k$
and they fail to intersect radical axis $x=0$ because the equation $y^2+k^2=0$ has no real solutions.