The midpoint of the chord of the circle x^2 + | vedclass

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(C) Let the midpoint of the chord be $M(h, k)$.Since the line $x - 2y = 2$ is the chord,the midpoint $M$ must lie on this line,so $h - 2k = 2$.The lin

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$\left( \frac{2}{5}, -\frac{4}{5} \right)$

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(C) Let the midpoint of the chord be $M(h, k)$.Since the line $x - 2y = 2$ is the chord,the midpoint $M$ must lie on this line,so $h - 2k = 2$.The line segment joining the center of the circle $(0, 0)$ to the midpoint $M(h, k)$ is perpendicular to the chord $x - 2y = 2$.The slope of the chord is $m_1 = \frac{1}{2}$.The slope of the line segment $OM$ is $m_2 = \frac{k - 0}{h - 0} = \frac{k}{h}$.Since $OM$ is perpendicular to the chord,$m_1 	imes m_2 = -1$,which gives $\frac{1}{2} 	imes \frac{k}{h} = -1$,so $k = -2h$.Substituting $k = -2h$ into the equation $h - 2k = 2$,we get $h - 2(-2h) = 2$,which implies $5h = 2$,so $h = \frac{2}{5}$.Then $k = -2 	imes \frac{2}{5} = -\frac{4}{5}$.Thus,the midpoint is $\left( \frac{2}{5}, -\frac{4}{5} ight)$.

The midpoint of the chord of the circle $x^2 + y^2 = 25$ intercepted by the line $x - 2y = 2$ is

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