The equation of the normal to the circle $x^{2} + y^{2} - 2 x = 0$ parallel to the line $x + 2 y = 3$ is

x + 2 y + 3 = 0

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x + 2 y - 3 = 0

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x + 2 y - 1 = 0

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x + 2 y + 1 = 0

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Solution

The correct option is C $x + 2 y - 1 = 0$
Given circle is $x^{2} + y^{2} - 2 x = 0$
Centre of the circle, $C = (1, 0)$
Any line parallel to the line $x + 2 y = 3$ is
$x + 2 y + k = 0$
As this is normal, so it passes through the centre $C (1, 0)$
$1 + 2 (0) + k = 0 \Rightarrow k = - 1$
Therefore, the equation of normal is
$x + 2 y - 1 = 0$

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