The locus of centre of a circle which passes through the origin and cuts off a length of $4$ units on the line $x = 3$ is

x^{2} + 6 y = 13

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y^{2} + 6 x = 13

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x^{2} + 6 y = 0

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y^{2} + 6 x = 0

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Solution

The correct option is B $y^{2} + 6 x = 13$
Let $C (h, k)$ be centre (locus point)

$∴$ radius, $r = O C = \sqrt{h^{2} + k^{2}}$
$C M = | h - 3 |$
We know that,
$A B = 2 \sqrt{C B^{2} - C M^{2}}$
$\Rightarrow 4 = 2 \sqrt{(h^{2} + k^{2}) - (h - 3)^{2}}$
$\Rightarrow 4 = (h^{2} + k^{2}) - (h^{2} - 6 h + 9)$
$\Rightarrow k^{2} + 6 h = 13$

$∴$ Locus of $C (h, k)$ is
$y^{2} + 6 x = 13$

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