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

x + 2y - 1 = 0

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x - 2y + 1 = 0

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x + 2y + 1 = 0

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x - 2y - 1 = 0

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Solution

The correct option is A x + 2y - 1 = 0
the center of the given circle is (1,0).
Any line parallel to x + 2y = 3 is given by x + 2y + k = 0
And we know that normal passes through the center of the circle
$⟹$ 1 + 2(0) + k = 0
$⟹$ k = -1
Hence, the equation of the normal will be : x + 2y - 1 = 0

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