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 $2 x - y + 3 = 0$

x + y - 7 = 0

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2 x - y - 5 = 0

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x + 2 y - 3 = 0

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2 x + y - 3 = 0

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Solution

The correct option is A $x + 2 y - 3 = 0$
Equation of the circle is $x^{2} + y^{2} + 2 x - 4 y - 11 = 0.$
We know that the standard equation of the circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .
Comparing the given equation with the standard equation, we get $g = 1, f = - 2$ and $c = - 11$
We also know that coordinates of the centre of the circle $(- g, - f) = (- 1, 2) .$
Since the diameter perpendicularly bisects the chords intercepted on $2 x - y + 3 = 0,$
Therefore the equation of the diameter is $x + 2 y + k = 0.$
We also know that the diameter passes through the centre $(- g . - f)$
Therefore, $(- 1) + 2 \times 2 + k = 0 \Rightarrow k = - 3.$
Thus, the equation of the diameter is $x + 2 y - 3 = 0.$

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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 - 4 y - 11 = 0

2 x - y + 3 = 0

Equation of the diameter of the circle $x^{2} + y^{2} - 2 x + 4 y = 0$ which passes through the origin is

x^{2} + y^{2} - 2 x + 4 y = 0

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