The equation of the family of circles having | vedclass

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(A) The radical axis of the two circles $S_1: x^2 + y^2 - 4 = 0$ and $S_2: x^2 + y^2 + 2x + 4y - 6 = 0$ is given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 4) -

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$x^2 + y^2 + \lambda x + 2\lambda y - (4 + \lambda) = 0$

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(A) The radical axis of the two circles $S_1: x^2 + y^2 - 4 = 0$ and $S_2: x^2 + y^2 + 2x + 4y - 6 = 0$ is given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 4) - (x^2 + y^2 + 2x + 4y - 6) = 0$$-2x - 4y + 2 = 0$$x + 2y - 1 = 0$.The family of circles having the same radical axis is given by $S_1 + \lambda L = 0$,where $L$ is the radical axis.$x^2 + y^2 - 4 + \lambda(x + 2y - 1) = 0$$x^2 + y^2 + \lambda x + 2\lambda y - (4 + \lambda) = 0$.

The equation of the family of circles having the same radical axis as the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 2x + 4y = 6$ is:

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