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If the two circles, $x^2+y^2+2 g_1 x+2 f_1 y=0$ and $x^2+y^2+2 g_2 x+2 f_2 y=0$ touches each other, then
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The correct answer is:
$\frac{f_1}{g_1}=\frac{f_2}{g_2}$
Since the 2 circles touch each other
$\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1 \pm \mathrm{r}_2$
$\begin{aligned} & \Rightarrow \sqrt{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2}=\sqrt{g_1^2+f_1^2} \pm \sqrt{g_2^2+f_2^2} \\ & -2 g_1 g_2-2 f_1 f_2= \pm 2\left(\sqrt{g_1^2+f_1^2}\right)\left(\sqrt{g_2^2+f_2^2}\right) \\ & \left(g_1 g_2\right)^2+\left(f_1 f_2\right)^2+2 g_1 g_2 f_1 f_2=\left(g_1 g_2\right)^2+\left(f_1 f_2\right)^2+\left(g_1 f_2\right. \\ & 2=\frac{g_1 f_2}{g_2 f_1}+\frac{g_2 f_1}{g_1 f_2} \\ & \text { Let } \frac{g_1 f_2}{g_2 f_1}=a \\ & \Rightarrow 2=a+\frac{1}{a} \Rightarrow a^2+-2 a+1=0 \\ & \rightarrow a=1 \rightarrow \frac{g_1 f_2}{g_2 f_1}=1\end{aligned}$
$\frac{f_1}{g_1}=\frac{f_2}{g_2}$
$\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1 \pm \mathrm{r}_2$
$\begin{aligned} & \Rightarrow \sqrt{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2}=\sqrt{g_1^2+f_1^2} \pm \sqrt{g_2^2+f_2^2} \\ & -2 g_1 g_2-2 f_1 f_2= \pm 2\left(\sqrt{g_1^2+f_1^2}\right)\left(\sqrt{g_2^2+f_2^2}\right) \\ & \left(g_1 g_2\right)^2+\left(f_1 f_2\right)^2+2 g_1 g_2 f_1 f_2=\left(g_1 g_2\right)^2+\left(f_1 f_2\right)^2+\left(g_1 f_2\right. \\ & 2=\frac{g_1 f_2}{g_2 f_1}+\frac{g_2 f_1}{g_1 f_2} \\ & \text { Let } \frac{g_1 f_2}{g_2 f_1}=a \\ & \Rightarrow 2=a+\frac{1}{a} \Rightarrow a^2+-2 a+1=0 \\ & \rightarrow a=1 \rightarrow \frac{g_1 f_2}{g_2 f_1}=1\end{aligned}$
$\frac{f_1}{g_1}=\frac{f_2}{g_2}$
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