If $\theta$ is the angle between the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-8x-12y+43=0$ then $\left|7\sec \theta -18\cos \theta \right|=$

Let

$S_1:x^2+y^2-8x-12y+43=0$

$S_2:x^2+y^2-2x-4y-4=0$

If $\theta$ is the angle between circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g^{\text{'}}x+2f^{\text{'}}y+c^{\text{'}}=0$, then $\cos \theta =\frac{c+c^{\text{'}}-2\left(g{g}^{\text{'}}+f{f}^{\text{'}}\right)}{2\sqrt{g^2+f^2-c}\sqrt{{\left(g^{\text{'}}\right)}^2+{\left(f^{\text{'}}\right)}^2-c^{\text{'}}}}$

$\Rightarrow \cos \theta =\frac{43-4-2\left(4+12\right)}{2\sqrt{1+4+4}\sqrt{16+36-43}}$

$\Rightarrow \cos \theta =\frac{7}{18}$

$\Rightarrow \sec \theta =\frac{18}{7}$

Therefore,

$\left|7\sec \theta -18\cos \theta \right|=18-7=11$