Find the equation of the chord of the circle | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 6x + 8y = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3$ and $f = 4$. The cent

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$2x + y - 7 = 0$

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(B) The equation of the circle is $x^2 + y^2 - 6x + 8y = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3$ and $f = 4$. The center of the circle is $C(-g, -f) = (3, -4)$. Let the point where the chord is bisected be $M(5, -3)$. The chord is perpendicular to the radius $CM$. The slope of the radius $CM$ is $m_{CM} = \frac{-3 - (-4)}{5 - 3} = \frac{1}{2}$. Since the chord is perpendicular to the radius,the slope of the chord $m = -\frac{1}{m_{CM}} = -2$. The equation of the chord passing through $M(5, -3)$ with slope $m = -2$ is given by $y - y_1 = m(x - x_1)$. $y - (-3) = -2(x - 5)$ $y + 3 = -2x + 10$ $2x + y - 7 = 0$.

Find the equation of the chord of the circle $x^2 + y^2 - 6x + 8y = 0$ which is bisected at the point $(5, -3)$.

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