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

\frac{9}{2}

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2 \sqrt{2}

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3 \sqrt{2}

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\frac{3}{2}

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Solution

The correct option is B $2 \sqrt{2}$
Eq'n of common chord $\Rightarrow 2 x + 1 = 0 ∥ y a x i s$
therefore length of common chord = diff. b/w $y$ co-ordinates of $P 01$
$x = - 1 / 2$ in $C_{1}$
$\frac{1}{4} + y^{2} + 3 y + 0 = 0$
$y^{2} + 3 y + \frac{1}{4} = 0$
$α + β = - 3$ | $α β = \frac{1}{4}$
$| α - β | = \sqrt{(α + β)^{2} - 4 α β} = \sqrt{9 - 1}$
$= (2)$

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x^{2} + y^{2} + 2 x + 3 y + 1 = 0

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

x^{2} + y^{2} + 2 x + 2 y + 1 = 0, x^{2} + y^{2} + 4 x + 3 y + 2 = 0

x^{2} + y^{2} + 2 x + 2 y + 1 = 0; x^{2} + y^{2} + 4 x + 3 y + 2 = 0

x^{2} + y^{2} = 12

x^{2} + y^{2} - 4 x + 3 y - 2 = 0

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x^{2} + y^{2} + 2 x - 3 y - 4 = 0

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