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

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

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(D) The equation of a family of circles passing through the intersection of a circle $S=0$ and a line $L=0$ is given by $S+\lambda L=0$.Here,$S = x^2+y^2-4x-6y-21=0$ and $L = 3x+4y+5=0$.So,the equation of the required circle is $(x^2+y^2-4x-6y-21) + \lambda(3x+4y+5) = 0$.Since this 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$$16\lambda = 32 \Rightarrow \lambda = 2$.Substituting $\lambda = 2$ back into the family equation:$(x^2+y^2-4x-6y-21) + 2(3x+4y+5) = 0$$x^2+y^2-4x-6y-21+6x+8y+10 = 0$$x^2+y^2+2x+2y-11 = 0$.Thus,the correct option is $D$.

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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