The angle between the circles $x^2+y^2+4x-14y+28=0$ and $x^2+y^2-12x-6y-4=0$ is

For the given circles we have $C_1=(-2,7),r_1=\sqrt{4+49-28}=5$;

$C_2=\left(6.3\right),r_2=\sqrt{36+9+4}=7,\text{ and }d=C_1{C}_2=\sqrt{64+16}=\sqrt{80}$

Let $\theta$ be the angle between the circles

$\cos \theta =\frac{d^2-r_1^2-r_2^2}{2r_1{r}_2}=\frac{80-25-49}{2\cdot 5\cdot 7}=\frac{6}{2\cdot 5\cdot 7}=\frac{3}{35}.\therefore \theta ={\mathrm{Cos}}^{-1}\left(\frac{3}{35}\right)$