If theta is the angle between the circles x^2 | vedclass

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(A) The given circles are $C_1: x^2+y^2-4x+2y-4=0$ and $C_2: x^2+y^2-2x+4y-11=0$.For $C_1$,the center $O_1 = (2, -1)$ and radius $r_1 = \sqrt{2^2 + (-

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$\frac{\sqrt{47}}{24}$

Vedclass · Answer

(A) The given circles are $C_1: x^2+y^2-4x+2y-4=0$ and $C_2: x^2+y^2-2x+4y-11=0$.For $C_1$,the center $O_1 = (2, -1)$ and radius $r_1 = \sqrt{2^2 + (-1)^2 - (-4)} = \sqrt{4+1+4} = 3$.For $C_2$,the center $O_2 = (1, -2)$ and radius $r_2 = \sqrt{1^2 + (-2)^2 - (-11)} = \sqrt{1+4+11} = 4$.The distance between centers $d = \sqrt{(2-1)^2 + (-1 - (-2))^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.The angle $	heta$ between two circles is given by $\cos 	heta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2}$.$\cos 	heta = \frac{2 - 9 - 16}{2(3)(4)} = \frac{-23}{24}$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - (\frac{-23}{24})^2 = 1 - \frac{529}{576} = \frac{47}{576}$.Therefore,$\sin 	heta = \sqrt{\frac{47}{576}} = \frac{\sqrt{47}}{24}$.

If $\theta$ is the angle between the circles $x^2+y^2-4x+2y-4=0$ and $x^2+y^2-2x+4y-11=0$,then $\sin \theta=$

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