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

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Solution

The correct option is A $2 \sqrt{g_{1}^{2} + f_{1}^{2} - c_{1}}$
Given $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ bisects the circumference of circle
$x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ that means.
common chord of this two circle is diameter of $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} =$
$∴$ length of common chord $= 2 \times \sqrt{g_{1}^{2} + f_{1}^{2} - c_{1}}$

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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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