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Find the equation of the normal to the circle `x^2+y^2-2x=0` parallel to the line `x+2y=3.`

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Find the equation of the normal to the circle `x^2+y^2-2x=0` parallel to the line `x+2y=3.`
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Slope of given line is `-(1)/(2)`.
So, slope of normal is `-(1)/(2)`.
Normal passes through the centre of the circle.
Here, centre is (1,0).
Therefore, equation of normal is
`y=0=-(1)/(2)(x-1)`
or `x+2y-1=0`
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