If the angle between the circles x^2+y^2-2x+2 | vedclass

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(B) The equations of the circles are $x^2+y^2-2x+2y+1=0$ and $x^2+y^2+2x-2y+k=0$. For the first circle,center $C_1 = (1, -1)$ and radius $r_1 = \sqrt{

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$k$ is an irrational number

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(B) The equations of the circles are $x^2+y^2-2x+2y+1=0$ and $x^2+y^2+2x-2y+k=0$. For the first circle,center $C_1 = (1, -1)$ and radius $r_1 = \sqrt{1^2+(-1)^2-1} = 1$. For the second circle,center $C_2 = (-1, 1)$ and radius $r_2 = \sqrt{(-1)^2+1^2-k} = \sqrt{2-k}$. The distance between centers $d = \sqrt{(1 - (-1))^2 + (-1 - 1)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$. The angle $	heta$ between the circles is given by $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$. Given $	heta = \frac{\pi}{3}$,so $\cos \frac{\pi}{3} = \frac{1}{2}$. Substituting the values: $\frac{1}{2} = \frac{1^2 + (\sqrt{2-k})^2 - (2\sqrt{2})^2}{2(1)(\sqrt{2-k})} = \frac{1 + 2 - k - 8}{2\sqrt{2-k}} = \frac{-5-k}{2\sqrt{2-k}}$. Thus,$\sqrt{2-k} = -5-k$. Squaring both sides: $2-k = (-5-k)^2 = 25 + k^2 + 10k$. $k^2 + 11k + 23 = 0$. Using the quadratic formula,$k = \frac{-11 \pm \sqrt{121 - 4(23)}}{2} = \frac{-11 \pm \sqrt{121 - 92}}{2} = \frac{-11 \pm \sqrt{29}}{2}$. Since $\sqrt{29}$ is irrational,$k$ is an irrational number.

If the angle between the circles $x^2+y^2-2x+2y+1=0$ and $x^2+y^2+2x-2y+k=0$ is $\frac{\pi}{3}$,then

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