The point (3,-4) lies on both the circles x^2 | vedclass

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(D) The given circles are $S_1: x^2+y^2-2x+8y+13=0$ and $S_2: x^2+y^2-4x+6y+11=0$. For $S_1$,the center $C_1 = (1, -4)$ and radius $r_1 = \sqrt{1^2+(-

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$135^{\circ}$

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(D) The given circles are $S_1: x^2+y^2-2x+8y+13=0$ and $S_2: x^2+y^2-4x+6y+11=0$. For $S_1$,the center $C_1 = (1, -4)$ and radius $r_1 = \sqrt{1^2+(-4)^2-13} = \sqrt{1+16-13} = 2$. For $S_2$,the center $C_2 = (2, -3)$ and radius $r_2 = \sqrt{2^2+(-3)^2-11} = \sqrt{4+9-11} = \sqrt{2}$. The distance between the centers $d = C_1C_2 = \sqrt{(2-1)^2+(-3-(-4))^2} = \sqrt{1^2+1^2} = \sqrt{2}$. The angle $	heta$ between the circles is given by $\cos 	heta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$. Substituting the values,$\cos 	heta = \frac{(\sqrt{2})^2 - 2^2 - (\sqrt{2})^2}{2 	imes 2 	imes \sqrt{2}} = \frac{2-4-2}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}}$. Thus,$	heta = 135^{\circ}$.

The point $(3,-4)$ lies on both the circles $x^2+y^2-2x+8y+13=0$ and $x^2+y^2-4x+6y+11=0$. Then,the angle between the circles is

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