The angle at which the circle $x^2+y^2+8x-2y-9=0,x^2+y^2-2x+8y-7=0$ intersect is

If $\theta$ is the angle b/w the circles $\cos \theta =\frac{d^2-r_1^2-r_2^2}{2r_1{r}_2}$
Here $\cos \theta =\frac{d^2-r_1^2-r_2^2}{2r_1{r}_2}$ 
$\cos \theta =\frac{d^2-r_1^2-r_2^2}{2r_1{r}_2},d=C_1{C}_2=\sqrt{50},r_1=\sqrt{26},r_2=\sqrt{24}$
$\therefore \cos \theta =\frac{50-26-24}{2.\sqrt{26}\sqrt{24}}=0\Rightarrow \theta =\frac{\pi }{2}$