If the line $x + 2 k y + 3 = 0$ is a diameter of the circle $x^{2} + y^{2} - 6 x + 2 y = 0$ , then $k$ is equal to

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Solution

The correct option is A $3$
Given equation of circle is
$x^{2} + y^{2} - 6 x + 2 y = 0$ , whose
centre $= (3, - 1)$ and radius $= \sqrt{9 + 1} = \sqrt{10}$
Clearly, the centre lies on the line
$x + 2 k y + 3 = 0$
We have, $3 - 2 k + 3 = 0$
$\Rightarrow 6 - 2 k = 0$
$∴ k = 3$

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