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Slope Point Form of a Line

The slope of ...

Question

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

A

1

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B

3

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C

5

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D

8

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Solution

The correct option is C $8$
$e q^{n}$ of radial axis is given by $s_{1} - s_{2} = 0$
$s_{1} = x^{2} + y^{2} + 3 x + 4 y - 5 = 0$
$s_{2} = x^{2} + y^{2} - 5 x + 5 y - 6 = 0$
$∴ s_{1} - s_{2} = 0$
$\Rightarrow 8 x - y + 1 = 0$
slope $= m = \frac{- 8}{- 1} = 8$

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Slope Point Form of a Line

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