The equation of the chord of the circle x^2+y | vedclass

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(D) The equation of a circle is given by $x^2+y^2-4x=0$.Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-2$,$f=0$,and $c=0$.The c

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$x=1$

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(D) The equation of a circle is given by $x^2+y^2-4x=0$.Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-2$,$f=0$,and $c=0$.The center of the circle is $(-g, -f) = (2, 0)$.The equation of a chord of a circle with a given midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T = xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c$ and $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.Here,$(x_1, y_1) = (1, 0)$.$T = x(1) + y(0) - 2(x+1) + 0(y+0) + 0 = x - 2x - 2 = -x - 2$.$S_1 = (1)^2 + (0)^2 - 4(1) = 1 - 4 = -3$.Equating $T = S_1$,we get $-x - 2 = -3$,which simplifies to $x = 1$.

The equation of the chord of the circle $x^2+y^2-4x=0$ whose midpoint is $(1,0)$ is

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