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

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { (a) } x^2+y^2+2 x-4 y+1=0 \ & 	herefore g_1=1, f_1=-2, c_1=1 \ & x^2+y^2-4 x-2 y+c=0 \ & 	herefore g_1=-2, f_2=-1, c_2=\mathrm{c} \end{aligned}$ Now, angle between the circles is $\begin{aligned} & \cos (	heta)=\frac{c_1+c_2-2\left(\mathrm{~g}_1 \mathrm{~g}_2+\mathrm{f}_1 \mathrm{f}_2ight)}{2 \sqrt{g_1^2+f_1^2-c_1} \sqrt{g_2^2+f_2^2-c_2}} \ & \Rightarrow \cos \left(\frac{\pi}{4}ight)=\frac{1+c-2(1 	imes(-2)+(-2)(-1))}{\sqrt[2]{1+4-1} \sqrt{4+1-c}} \ & \Rightarrow \frac{1}{\sqrt{2}}=\frac{1+c}{4 \sqrt{5-c}} \Rightarrow 16(5-c)=2(1+c)^2 \ & \Rightarrow c^2+10 c-39=0 \Rightarrow(c-3)(c+13)=0 \ & \Rightarrow c=3 	ext { or } c=-13 \end{aligned}$

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

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