Explanation

(d)
Given circles are $x^{2} + y^{2} - 2x + 8y + 13 = 0$ and $x^{2} + y^{2} - 4x + 6y + 11 = 0$
Here, $C_{1} = (1,\ - 4),\ C_{2} = (2,\ - 3)$
$\Rightarrow \ \ \ r_{1} = \sqrt{1 + 16 - 13} = 2$
And $r_{2} = \sqrt{4 + 9 - 11} = \sqrt{2}$
Now, $d = C_{1}C_{2} = \sqrt{(2 - 1)^{2} + ( - 3 + 4)^{2}} = \sqrt{2}$
$\therefore\ \cos\theta = \frac{{|d}^{2} - r_{1}^{2} - r_{2}^{2}|}{2r_{1}r_{2}} = \frac{|2 - 4 - 2|}{2 \times 2 \times \sqrt{2}} = \frac{1}{\sqrt{2}}$
$\Rightarrow \ \theta = 45{^\circ}$