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

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(A) The given circle is $x^2+y^2-2x=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$,$f=0$,and $c=0$. The center of th

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$x+2y-1=0$

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(A) The given circle is $x^2+y^2-2x=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$,$f=0$,and $c=0$. The center of the circle is $(-g, -f) = (1, 0)$. Any normal to a circle always passes through its center. The normal is parallel to the line $x+2y-3=0$. The slope of the line $x+2y-3=0$ is $m = -\frac{1}{2}$. Since the normal is parallel to this line,the slope of the normal is also $m = -\frac{1}{2}$. The equation of the line passing through $(1, 0)$ with slope $m = -\frac{1}{2}$ is given by $y - y_1 = m(x - x_1)$. Substituting the values: $y - 0 = -\frac{1}{2}(x - 1)$. $2y = -x + 1$,which simplifies to $x + 2y - 1 = 0$.

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

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