The angle at which the circles (x - 1)^2 + y^ | vedclass

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(B) The centers of the circles are $C_1 = (1, 0)$ and $C_2 = (0, 2)$,and their radii are $r_1 = \sqrt{10}$ and $r_2 = \sqrt{5}$.The distance between t

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$\frac{\pi}{4}$

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(B) The centers of the circles are $C_1 = (1, 0)$ and $C_2 = (0, 2)$,and their radii are $r_1 = \sqrt{10}$ and $r_2 = \sqrt{5}$.The distance between the centers $d$ is given by $d = \sqrt{(1 - 0)^2 + (0 - 2)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.The angle of intersection $	heta$ between two circles is given by $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2 r_1 r_2}$.Substituting the values,we get $\cos 	heta = \frac{10 + 5 - 5}{2 \cdot \sqrt{10} \cdot \sqrt{5}} = \frac{10}{2 \cdot \sqrt{50}} = \frac{10}{2 \cdot 5\sqrt{2}} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = \frac{\pi}{4}$.

The angle at which the circles $(x - 1)^2 + y^2 = 10$ and $x^2 + (y - 2)^2 = 5$ intersect is

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