The equation of that diameter of the circle $x^{2} + y^{2} - 6 x + 2 y - 8 = 0$ which passes through the origin is

6 x - y = 0

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3 x + 2 y = 0

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x + 3 y = 0

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3 x - y = 0

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Solution

The correct option is C $x + 3 y = 0$
Centre of the circle;
$C = (3, - 1)$
The equation of the diameter joining the origin $(0, 0)$ and $C = (3, - 1)$ is,

$y - 0 = \frac{- 1 - 0}{3 - 0} (x - 0) \Rightarrow 3 y + x = 0$

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

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} - 6 x + 2 y - 8 = 0

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