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

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$y^{2} + 6 x = 0$

$y^{2} + 6 x = 13$

$y^{2} + 6 x = 10$

$x^{2} + 6 y = 13$

answer is B.

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Detailed Solution

Let $(x_{1}, y_{1})$ be the centre of the circle.

Circle cuts off a length of 4 units from the line $x = 3$

$\Rightarrow$ Circle cuts the line at $(3, β), (3, β + 4) .$

Circle passes through the origin $\Rightarrow$ Equation of the circle is $x^{2} + y^{2} - 2 x_{1} x - 2 y_{1} y = 0 .$

$(3, β), (3, β + 4)$ lie on the circle $\Rightarrow 9 + β^{2} - 6 x_{1} - 2 y_{1} β = 0 \to (1),$

$\begin{array}{l} 9 + (β + 4)^{2} - 6 x_{1} - 2 y_{1} (β + 4) = 0 \to (2) \\ (2) - (1) \Rightarrow 8 β + 16 - 8 y_{1} = 0 \Rightarrow y_{1} = \bar{β} + \bar{2} \Rightarrow β = y_{1} - 2 \\ (1) \Rightarrow 9 + {(y_{1} - 2)}^{2} - 6 x_{1} - 2 y_{1} (y_{1} - 2) = 0 \Rightarrow 9 + y_{1}^{2} + 4 - 4 y_{1} - 6 x_{1} - 2 y_{1}^{2} + 4 y_{1} = 0 \end{array}$