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

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$x^2 + y^2 + 2x + 2y - 11 = 0$

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(B) The family of circles passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$,where $\lambda \in \mathbb{R}$.For the given circle $x^2 + y^2 - 4x - 6y - 21 = 0$ and line $3x + 4y + 5 = 0$,the equation of the family of circles is:$x^2 + y^2 - 4x - 6y - 21 + \lambda(3x + 4y + 5) = 0$Since the circle passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ into the equation:$(1)^2 + (2)^2 - 4(1) - 6(2) - 21 + \lambda(3(1) + 4(2) + 5) = 0$$1 + 4 - 4 - 12 - 21 + \lambda(3 + 8 + 5) = 0$$-32 + 16\lambda = 0$$16\lambda = 32$$\lambda = 2$Substituting $\lambda = 2$ back into the family equation:$x^2 + y^2 - 4x - 6y - 21 + 2(3x + 4y + 5) = 0$$x^2 + y^2 - 4x - 6y - 21 + 6x + 8y + 10 = 0$$x^2 + y^2 + 2x + 2y - 11 = 0$

Find the equation of the circle passing through the intersection of the circle $x^2 + y^2 - 4x - 6y - 21 = 0$ and the line $3x + 4y + 5 = 0$,and also passing through the point $(1, 2)$.

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