One of the diameters, of the circle $x^{2} + y^{2} - 12 x + 4 y + 6 = 0$ is given by

x + y = 0

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x + 3 y = 0

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x = y

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3 x + 2 y = 0

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Solution

The correct option is A $x + 3 y = 0$
Give equation of circle is $x^{2} + y^{2} - 12 x + 4 y + 6 = 0$ whose centre is $(6, - 2)$ .
We know that centre of the circle lies on the diameter of the circle.
Out of all options
$\Rightarrow$ Point $(6, - 2)$ satisfy the equation $x + 3 y = 0$
i.e., $6 + 3 \times (- 2) = 0 \Rightarrow 0 = 0$
Hence, option (b) is correct.

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