The radical axis of circles $x^{2} + y^{2} + 3 x + 4 y - 5 = 0$ and $x^{2} + y^{2} - 5 x + 5 y - 6 = 0$ is

8 x + y + 1 = 0

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8 x - y + 1 = 0

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8 x - 8 y + 1 = 0

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- 8 x + y + 1 = 0

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Solution

The correct option is D $8 x - y + 1 = 0$
$e q^{n}$ of radical axis is given by $s_{1} - s_{2} = 0$
$s_{1} = x^{2} + y^{2} + 3 x + 4 y - 5 = 0$
$s_{2} = x^{2} + y^{2} - 5 x + 5 y - 6 = 0$
$= s_{1} - s_{2} = 0$
$\Rightarrow 8 x - y + 1 = 0$

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