The equation of the common radical axis of the orthogonal coaxial system of the coaxial system $(x^{2} + y^{2} - 4 x - 6 y + 5) + λ (2 x + 3 y + 4) = 0$ is

2 x + 3 y + 4 = 0

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3 x - 2 y + 4 = 0

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2 x + 3 y - 4 = 0

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3 x - 2 y - 1 = 0

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Solution

The correct option is B $2 x + 3 y + 4 = 0$
$(x^{2} + y^{2} - 4 x - 6 y + 5) + λ (2 x + 3 y + 4) = 0$ represent a family of circles Then for $λ = 0, s_{1}; x^{2} + y^{2} - 4 x - 6 y + 5 = 0$ ---- (1)
For $λ = 1 s^{2}; x^{2} + y^{2} - 2 x - 3 y + 9 = 0$ ---(2)
Then radical axis of $s_{1} a n d s_{2}$ is given by
$s_{1} - s_{2} = 0$
$- 2 x - 3 y - 4 = 0$

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