If the angle between the circles x^2+y^2+4x-5 | vedclass

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(A) The equations of the circles are $x^2+y^2+4x-5=0$ and $x^2+y^2+2\lambda y-4=0$. Comparing these with the general equation $x^2+y^2+2gx+2fy+c=0$,we

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$\pm \sqrt{5}$

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(A) The equations of the circles are $x^2+y^2+4x-5=0$ and $x^2+y^2+2\lambda y-4=0$. Comparing these with the general equation $x^2+y^2+2gx+2fy+c=0$,we get: For the first circle: $g_1=2, f_1=0, c_1=-5$. For the second circle: $g_2=0, f_2=\lambda, c_2=-4$. The angle $	heta$ between two circles is given by $\cos 	heta = \frac{2g_1g_2+2f_1f_2-c_1-c_2}{2\sqrt{g_1^2+f_1^2-c_1}\sqrt{g_2^2+f_2^2-c_2}}$. Given $	heta = \frac{\pi}{3}$,so $\cos \frac{\pi}{3} = \frac{1}{2}$. Substituting the values: $\frac{1}{2} = \frac{2(2)(0) + 2(0)(\lambda) - (-5) - (-4)}{2\sqrt{2^2+0^2-(-5)}\sqrt{0^2+\lambda^2-(-4)}}$. $\frac{1}{2} = \frac{0+0+5+4}{2\sqrt{4+5}\sqrt{\lambda^2+4}} = \frac{9}{2\sqrt{9}\sqrt{\lambda^2+4}}$. $\frac{1}{2} = \frac{9}{2(3)\sqrt{\lambda^2+4}} = \frac{3}{2\sqrt{\lambda^2+4}}$. $\sqrt{\lambda^2+4} = 3$. Squaring both sides: $\lambda^2+4 = 9$. $\lambda^2 = 5$,which implies $\lambda = \pm \sqrt{5}$.

If the angle between the circles $x^2+y^2+4x-5=0$ and $x^2+y^2+2\lambda y-4=0$ is $\frac{\pi}{3}$,then $\lambda=$

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