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(B) The equation of a family of circles passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$.Given

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${x^2} + {y^2} + 4x - 2y - 4 = 0$

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(B) The equation of a family of circles passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$.Given $S = {x^2} + {y^2} - 2x - 6y + 6 = 0$ and $L = 3x + 2y - 5 = 0$,the equation is:$({x^2} + {y^2} - 2x - 6y + 6) + \lambda (3x + 2y - 5) = 0$Since the circle passes through the point $(-2, 4)$,we substitute $x = -2$ and $y = 4$ into the equation:$((-2)^2 + (4)^2 - 2(-2) - 6(4) + 6) + \lambda (3(-2) + 2(4) - 5) = 0$$(4 + 16 + 4 - 24 + 6) + \lambda (-6 + 8 - 5) = 0$$6 + \lambda (-3) = 0$$3\lambda = 6 \implies \lambda = 2$Substituting $\lambda = 2$ back into the family equation:$({x^2} + {y^2} - 2x - 6y + 6) + 2(3x + 2y - 5) = 0$${x^2} + {y^2} - 2x - 6y + 6 + 6x + 4y - 10 = 0$${x^2} + {y^2} + 4x - 2y - 4 = 0$

The equation of the circle passing through the point $(-2, 4)$ and through the points of intersection of the circle ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and the line $3x + 2y - 5 = 0$ is:

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