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If the equation of the circle whose one diameter is the common chord of the circles $x^{2} + y^{2} + 4 x + 2 y - 4 = 0$ and $x^{2} + y^{2} - 2 x - 4 y - 4 = 0$ is $x^{2} + y^{2} + a x + b y + c = 0$ then $a + b - c =$

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Detailed Solution

The common chord is $S - S' = 0$
Equation of common chord is $S + λ L = 0$

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