The length of the common chord of circles $x^{2} + y^{2} - 6 x - 16 = 0$ and $x^{2} + y^{2} - 8 y - 9 = 0$ is-

10 \sqrt{3}

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5 \sqrt{3}

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5 \sqrt{3} / 2

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N o n e o f t h e s e

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Solution

The correct option is B $5 \sqrt{3}$
$\begin{matrix} x^{2} + y^{2} - 6 x - 16 = 0 x^{2} + y^{2} - 8 y - 9 = 0 e q u a t i o n o f c o m m o n c h o r d : - 6 x - 16 + 8 y + 9 = 0 8 y - 6 x - 7 = 0 8 y = 6 x + 7 L e n g t h o f p e r p e n d i c u l a r f r o m (3, 0) o n c o m m o n c h o r d : \frac{| - 18 - 7 |}{\sqrt{36 + 64}} = \frac{25}{10} = \frac{5}{2} L e n g t h o f c o m m o n c h o r d = 2 \sqrt{{(5)}^{2} - {(\frac{5}{2})}^{2}} = 2 \times 5 \sqrt{1 - \frac{1}{4}} = 2 \times 5 \sqrt{\frac{3}{4}} = 5 \sqrt{3} \end{matrix}$

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x^{2} + y^{2} + 4 y - 5 = 0

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The common chord of the circle $x^{2} + y^{2} + 4 x + 1 = 0$ and $x^{2} + y^{2} + 6 x + 2 y + 3 = 0$ is

x^{2} + y^{2} - 6 x - 4 y + 9 = 0

x^{2} + y^{2} - 8 x - 6 y + 23 = 0

x^{2} + y^{2} + x + 2 y + 3 = 0, x^{2} + y^{2} + 2 x + 4 y + 5 = 0,

x^{2} + y^{2} - 7 x - 8 y - 9 = 0.

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