The equation of a chord of the circle $x^{2} + y^{2} - 4 x = 0$ which is bisected at the point (1,1) is

x + y =2

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3x - y =2

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x -2y +1 =0

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x - y = 0

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Solution

The correct option is D x - y = 0
The chord with given mid-point is given by $S_{1} = T$
Here $T = x . 1 - 2 (x + 1) = 0$ or $y - x - 2 = 0$ and $S_{1} = (1)^{2} + (1)^{2} - 4 (1) = - 2$
Therefore Chord is $y - x - 2 = - 2 o r x - y = 0$

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The common chord of the circle $x^{2} + y^{2} + 4 x + 1 = 0$ and $x^{2} + y^{2} + 6 x + 2 y + 3 = 0$ is

x^{2} + y^{2} - 2 x - 2 y - 2 = 0

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