The equation to the chord of the circle $x^{2} + y^{2} = 16$ which is bisected at $(2, - 1)$ is?

2 x + y = 16

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2 x - y = 16

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x + 2 y = 5

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2 x - y = 5

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Solution

The correct option is D $2 x - y = 16$
$x^{2} + y^{2} = 16$ bisected at $(2, - 1)$
Then equation of chord $x x_{1} + y y_{1} = 16 o r T = 0 2 x - y = 16$

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Similar questions

The equation of the diameter of the circle $x^{2} + y^{2} + 2 x - 4 y - 11 = 0$ which bisects the chords intercepted on the line $2 x - y + 3 = 0$ is

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

120^{0}

3 x + y + 5 = 0

x^{2} + y^{2} = 16

3 x + y + 5 = 0

x^{2} + y^{2} = 16

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2 x + y = 5

x^{2} + y^{2} - 6 x + 2 y = 0

2 x + y = 5

x^{2} + y^{2} - 6 x + 2 y = 0.

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