For the first circle $S_1: x^2+y^2=25$: its center is $C_1=(0,0)$ and its radius is $r_1=5$. For the second circle $S_2: x^2+y^2-8x-6y+21=0$: its center is $C_2=(-(-8)/2, -(-6)/2)=(4,3)$. Its radius is $r_2 = \sqrt{4^2+3^2-21} = \sqrt{16+9-21} = \sqrt{25-21} = \sqrt{4}=2$. The distance between the centers $d = C_1C_2 = \sqrt{(4-0)^2+(3-0)^2} = \sqrt{16+9} = \sqrt{25}=5$. To check if the circles intersect, we compare $d$ with $r_1+r_2$ and $|r_1-r_2|$. $r_1+r_2 = 5+2=7$. $|r_1-r_2| = |5-2|=3$. Since $|r_1-r_2| < d < r_1+r_2$ (i.e., $3 < 5 < 7$), the circles intersect. The angle $\theta$ between two intersecting circles is given by the formula $\cos\theta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2}$. Substituting the values: $\cos\theta = \frac{5^2+2^2-5^2}{2(5)(2)} = \frac{25+4-25}{20} = \frac{4}{20} = \frac{1}{5}$. Therefore, the angle between the circles is $\theta = \cos^{-1}\left(\frac{1}{5}\right)$.