If the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ bisects the circumference of the circle $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ then the length of the common chord of the circles is

2 \sqrt{{g_{1}}^{2} + {f_{1}}^{2} - c_{1}}

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\sqrt{{g_{1}}^{2} + {f_{1}}^{2} - c_{1}}

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\sqrt{g^{2} + f^{2} - c}

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\sqrt[2]{g^{2} + f^{2} - c}

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Solution

The correct option is A $2 \sqrt{{g_{1}}^{2} + {f_{1}}^{2} - c_{1}}$
$C_{2} : x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

$C_{1} : x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$

If $C_{2}$ bisects circumstances of $C_{1}$

then common chord $A B =$ Diameter of $C_{1}$

$= 2 \sqrt{g_{1}^{2} + f_{1}^{2} - c_{2}}$
$A$ is correct.

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Similar questions

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0

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

x^{2} + y^{2} + 2 g x + 2 f y = 0

x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0

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