The length of the common chord of the circles: $x^{2} + y^{2} + 6 x + 5 = 0$ and $x^{2} + y^{2} + 4 y - 5 = 0$ is

\sqrt{\frac{12}{13}}

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\frac{12}{\sqrt{13}}

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\frac{\sqrt{12}}{13}

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\sqrt{\frac{13}{12}}

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Solution

The correct option is C $\frac{12}{\sqrt{13}}$
$S_{1} : x^{2} + y^{2} + 6 x + 5 = 0$
$C_{1} \equiv (- 3, 0)$ and $r = 2$
$S_{2} : x^{2} + y^{2} + 4 y - 5 = 0$
$C_{2} \equiv (0, - 2)$ and $r = 3$
$e q^{n}$ of chord is $S_{1} - S_{2} = 0$
line $A B : 6 x - 4 y + 10 = 0 - - - - - (1)$
So, $C_{1} P = ∣ ∣ ∣ \frac{- 18 + 10}{\sqrt{52}} ∣ ∣ ∣$
$C_{1} P = \frac{4}{\sqrt{13}}$
So, In $Δ C_{1} A P,$
$A P = \sqrt{4 - \frac{16}{13}}$
$= \frac{6}{\sqrt{13}}$
So, length of chord is
$2 A P = \frac{12}{\sqrt{13}}$

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