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Equation of Radical Axis

lf 1, 2 is ...

Question

lf $(1, 2)$ is a limiting point of a coaxial system of circles containing the circle $x^{2} + y^{2} + x - 5 y + 9 = 0$ , then the equation of radical axis is

x + 3 y + 9 = 0

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3 x - y + 4 = 0

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x + 9 y - 4 = 0

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3 x - y - 1 = 0

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Solution

The correct option is D $3 x - y - 1 = 0$
Slope of circle
$2 x + 2 y y^{'} + 1 - 5 y^{'} = 0 y^{'} (2 y - 5) = - (2 x + 1) y^{'} = \frac{- (2 x + 1)}{(2 y - 5)}$
Slope at $(1, 2) y^{'} (1, 2) = \frac{- (2 (1) + 1)}{(2 (2) - 5)} = 3$
Equation of radial axis $(y - y_{1}) = m (x - x_{1}) (y - 2) = 3 (x - 1) 3 x - y - 1 = 0$

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