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

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(D) 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_

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$135^{\circ}$

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(D) 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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