A straight line passes through a fixed point $(2, 3)$ . Locus of the foot of the perpendicular on it drawn from the origin is

x^{2} + y^{2} + 2 x - 3 y = 0

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x^{2} + y^{2} - 2 x - 3 y = 0

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x^{2} + y^{2} - 2 x + 3 y = 0

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x^{2} + y^{2} + 2 x + 3 y = 0

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Solution

The correct option is B $x^{2} + y^{2} - 2 x - 3 y = 0$
(Slope of OA) $\cdot$ (Slope of AB) $= - 1$ $(∵ O A ⊥ A B)$
$(\frac{k - 0}{h - 0}) (\frac{k - 3}{h - 2}) = - 1$
$(k) (k - 3) + h (h - 2) = 0$
$k^{2} - 3 k + h^{2} - 2 = 0$
$x^{2} - 2 x + y^{2} - 3 y = 0$
$x^{2} + y^{2} - 2 x - 3 y = 0$ .

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