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If the circles $x^{2}+y^{2}+2 g x+2 f y=0$ and $x^{2}+y^{2}+2 g^{\prime} x+2 f^{\prime} y=0$ touch each other, then
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Verified Answer
The correct answer is:
$f^{\prime} g=f g^{\prime}$
For given condition,
$$
\begin{aligned}
\frac{2 g}{2 g^{\prime}} &=\frac{2 f}{2 f^{\prime}} \\
\Rightarrow \quad f^{\prime} &=f g^{\prime}
\end{aligned}
$$
$$
\begin{aligned}
\frac{2 g}{2 g^{\prime}} &=\frac{2 f}{2 f^{\prime}} \\
\Rightarrow \quad f^{\prime} &=f g^{\prime}
\end{aligned}
$$
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