The equation of the normal to the circle x^2 | vedclass

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(C) The given circle is $x^2 + y^2 - 2x = 0$. Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = 0$. The c

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$x + 2y - 1 = 0$

Vedclass · Answer

(C) The given circle is $x^2 + y^2 - 2x = 0$. Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$ and $f = 0$. The centre of the circle is $(-g, -f) = (1, 0)$. Any line parallel to $x + 2y = 3$ is of the form $x + 2y + \lambda = 0$. Since the normal to a circle always passes through its centre,the line $x + 2y + \lambda = 0$ must pass through $(1, 0)$. Substituting the point $(1, 0)$ into the equation: $1 + 2(0) + \lambda = 0$,which gives $\lambda = -1$. Thus,the equation of the normal is $x + 2y - 1 = 0$.

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

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