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

2 x + y - 7 = 0

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x + 2 y + 1 = 0

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2 x - y - 13 = 0

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x - 2 y - 11 = 0

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Solution

The correct option is C $2 x + y - 7 = 0$
$\begin{matrix} x^{2} + y^{2} - 6 x + 8 y = 0 T = S_{1} 5 x - 3 y - 6 \frac{(x + 5)}{2} + \frac{8 (y - 3)}{2} = 25 + 9 - 6 \times 5 + 8 (- 3) 5 x - 3 y - 3 (x + 5) + 4 (y - 3) = 25 + 9 - 30 - 24 2 x + y - 27 = - 20 2 x + y = 7 O p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$

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2 x + y = 5

x^{2} + y^{2} - 6 x + 2 y = 0

2 x + y = 5

x^{2} + y^{2} - 6 x + 2 y = 0.

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