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

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(C) For the circle $x^2+y^2+2x-4y+1=0$,we have $g_1=1, f_1=-2, c_1=1$. For the circle $x^2+y^2-4x-2y+c=0$,we have $g_2=-2, f_2=-1, c_2=c$. The angle $

Vedclass · Accepted Answer

$3$ or $-13$

Vedclass · Answer

(C) For the circle $x^2+y^2+2x-4y+1=0$,we have $g_1=1, f_1=-2, c_1=1$. For the circle $x^2+y^2-4x-2y+c=0$,we have $g_2=-2, f_2=-1, c_2=c$. The angle $	heta$ between two circles is given by $\cos(	heta) = \frac{c_1+c_2-2(g_1g_2+f_1f_2)}{2\sqrt{g_1^2+f_1^2-c_1}\sqrt{g_2^2+f_2^2-c_2}}$. Given $	heta = \frac{\pi}{4}$,so $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Substituting the values: $\frac{1}{\sqrt{2}} = \frac{1+c-2((1)(-2)+(-2)(-1))}{2\sqrt{1^2+(-2)^2-1}\sqrt{(-2)^2+(-1)^2-c}}$. $\frac{1}{\sqrt{2}} = \frac{1+c-2(-2+2)}{2\sqrt{4}\sqrt{5-c}} = \frac{1+c}{4\sqrt{5-c}}$. Squaring both sides: $\frac{1}{2} = \frac{(1+c)^2}{16(5-c)}$. $8(5-c) = (1+c)^2 \Rightarrow 40-8c = 1+2c+c^2$. $c^2+10c-39=0$. $(c+13)(c-3)=0$. Thus,$c=3$ or $c=-13$.

If the angle between the circles $x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y+c=0$ is $\frac{\pi}{4}$,then $c=$

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