The slope of the radical axis of the circles $(x + 2)^{2} + (y + 3)^{2} = 25$ and
$(x + 1)^{2} + (y - 1)^{2} = 25$ is

- \frac{1}{4}

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\frac{1}{4}

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- 4

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- \frac{1}{2}

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Solution

The correct option is B $- \frac{1}{4}$
$e q^{n}$ of radial axis is $s_{1} - s_{2} = 0$
$∴ ((x + 2)^{2} + (y + 3)^{2} - 25) - ((x + 1)^{2} + (y - 1)^{2} - 25) = 0$
$2 x + 8 y - 11 = 0$
slope of radial axis $= \frac{- 2}{8} = - \frac{1}{4}$

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