Find the angle of intersection of the two cir | vedclass

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(D) For the first circle $x^2 + y^2 - 2x - 2y = 0$,the center $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2 + 1^2} = \sqrt{2}$.For the second circle $x^2

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$45$

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(D) For the first circle $x^2 + y^2 - 2x - 2y = 0$,the center $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2 + 1^2} = \sqrt{2}$.For the second circle $x^2 + y^2 = 4$,the center $C_2 = (0, 0)$ and radius $r_2 = 2$.The distance between the centers $d = C_1C_2 = \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{2}$.If $	heta$ is the angle of intersection,then $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$.Substituting the values,$\cos 	heta = \frac{2 + 4 - 2}{2 	imes \sqrt{2} 	imes 2} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$	heta = 45^\circ$.

Find the angle of intersection of the two circles $x^2 + y^2 - 2x - 2y = 0$ and $x^2 + y^2 = 4$ in degrees.

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