Find the length of the intercept made on the $Y$ axis by the circle $\;$ $x^2 + y^2 -2ax \sec \theta - 2by \tan \theta - b^2 = 0$
Equation of given circle is $\;\;$ $x^2 + y^2 -2ax \sec \theta - 2by \tan \theta - b^2 = 0$
When the circle cuts the $Y$ axis, its $x$ coordinate equals $0$.
Putting $x = 0$ in the equation of the given circle, we have,
$y^2 - 2by \tan \theta - b^2 = 0$
Solving for $y$, we get,
$\begin{aligned}
y & = \dfrac{2b \tan \theta \pm \sqrt{\left(-2b \tan \theta\right)^2 - 4 \times 1 \times \left(-b^2\right)}}{2} \\\\
& = \dfrac{2 b \tan \theta \pm \sqrt{4b^2 \tan^2 \theta + 4b^2}}{2} \\\\
& = \dfrac{2b \tan \theta \pm \sqrt{4b^2 \left(1 + \tan^2 \theta\right)}}{2} \\\\
& = \dfrac{2 b \tan \theta \pm 2 b \sec \theta}{2} \\\\
& = b \tan \theta \pm b \sec \theta
\end{aligned}$
$\therefore \;$ The given circle cuts the $Y$ axis at the points $\;$ $A \left(0, b \tan \theta + b \sec \theta\right)$ $\;$ and $\;$ $B \left(0, b \tan \theta - b \sec \theta\right)$
$\therefore \;$ Length of intercept made on the $Y$ axis by the given circle is
$= AB = \sqrt{\left(b \tan \theta + b \sec \theta - b \tan \theta + b \sec \theta\right)^2}$
i.e. $\;$ $AB = 2b \sec \theta$