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(D) The equation of a circle passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$. Here,$S = x^2+y^

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

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(D) The equation of a circle passing through the intersection of a circle $S = 0$ and a line $L = 0$ is given by $S + \lambda L = 0$. Here,$S = x^2+y^2-4x-6y-21$ and $L = 3x+4y+5$. So,the equation is $(x^2+y^2-4x-6y-21) + \lambda(3x+4y+5) = 0$. Since the circle passes through $(1,2)$,we substitute $x=1$ and $y=2$: $(1^2+2^2-4(1)-6(2)-21) + \lambda(3(1)+4(2)+5) = 0$. $(1+4-4-12-21) + \lambda(3+8+5) = 0$. $-32 + 16\lambda = 0 \implies 16\lambda = 32 \implies \lambda = 2$. Substituting $\lambda = 2$ back into the equation: $(x^2+y^2-4x-6y-21) + 2(3x+4y+5) = 0$. $x^2+y^2-4x-6y-21 + 6x+8y+10 = 0$. $x^2+y^2+2x+2y-11 = 0$.

The equation of the circle passing through the point $(1,2)$ and through the points of intersection of $x^2+y^2-4x-6y-21=0$ and $3x+4y+5=0$ is given by

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