If the two circles $x^{2} + y^{2} + 2 g x + 2 f y = 0$ and $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0$ touch each other, then

f_{1} g = f g_{1}

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f f_{1} = g g_{1}

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f^{2} + g^{2} = f_{1}^{2} + g_{1}^{2}

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None of these

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Solution

The correct option is A $f_{1} g = f g_{1}$
$C_{1} (- g, - f)$ and $r_{1} = \sqrt{g^{2} + f^{2}}$
$C_{2} (- g_{1}, - f_{1})$ and $r_{2} = \sqrt{{g_{1}}^{2} + {f_{1}}^{2}}$
Since, circles touch each other
$C_{1} C_{2} = r_{1} + r_{2}$
$\sqrt{{(g - g_{1})}^{2} + {(f - f_{1})}^{2}} = \sqrt{g^{2} + f^{2}} + \sqrt{g_{1}^{2} + {f_{1}}^{2}}$
$\Rightarrow g^{2} + {g_{1}}^{2} - 2 g g_{1} + f^{2} + {f_{1}}^{2} - 2 f f_{1} = g^{2} + f^{2} + g_{1}^{2} + {f_{1}}^{2} + 2 \sqrt{g^{2} + f^{2}} \sqrt{g_{1}^{2} + {f_{1}}^{2}}$
$\Rightarrow - 2 (g g_{1} + f f_{1}) = 2 \sqrt{g^{2} {g_{1}}^{2} + f^{2} {f_{1}}^{2} + g^{2} {f_{1}}^{2} + f^{2} {g_{1}}^{2}}$
$\Rightarrow g^{2} {g_{1}}^{2} + f^{2} {f_{1}}^{2} + 2 g g_{1} f f_{1} = g^{2} {g_{1}}^{2} + f^{2} {f_{1}}^{2} + g^{2} {f_{1}}^{2} + f^{2} {g_{1}}^{2}$
$\Rightarrow g^{2} {f_{1}}^{2} + f^{2} {g_{1}}^{2} - 2 g g_{1} f f_{1} = 0$
$\Rightarrow g f_{1} - g_{1} f = 0$

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