The length of the common chord of the circles $x^{2} + y^{2} = 12$ and $x^{2} + y^{2} - 4 x + 3 y - 2 = 0$ , is

4 \sqrt{2}

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5 \sqrt{2}

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2 \sqrt{2}

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6 \sqrt{2}

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Solution

The correct option is A $4 \sqrt{2}$
$\begin{matrix} x^{2} + y^{2} = 12 x^{2} + y^{2} - 4 x + 3 y - 2 = 0 x^{2} + y^{2} - 12 = 0 {(x - 2)}^{2} + {(y + \frac{3}{2})}^{2} = 2 + 4 + \frac{9}{4} - 4 x + 3 y - 2 + 12 = 0 {(x - 2)}^{2} + {(y + \frac{3}{2})}^{2} = \frac{33}{4} - 4 x + 3 y + 10 = 0 4 x - 3 y - 10 = 0 \end{matrix}$
Length of perpendicular from $(0, 0) = \frac{| - 10 |}{\sqrt{16 + 9}} = \frac{10}{5} = 2$
Length of common chord $= 2 \sqrt{12 - 4} = 2 \times 2 \sqrt{2} = 4 \sqrt{2}$

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