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(D) Let the equation of the family of circles passing through the intersection of $S_{1} = x^{2}+y^{2}-6x=0$ and $S_{2} = x^{2}+y^{2}-4y=0$ be $S_{1}

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$(-3, 6)$

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(D) Let the equation of the family of circles passing through the intersection of $S_{1} = x^{2}+y^{2}-6x=0$ and $S_{2} = x^{2}+y^{2}-4y=0$ be $S_{1} + \lambda S_{2} = 0$ for $\lambda 
eq -1$.$(x^{2}+y^{2}-6x) + \lambda(x^{2}+y^{2}-4y) = 0$$(1+\lambda)x^{2} + (1+\lambda)y^{2} - 6x - 4\lambda y = 0$Dividing by $(1+\lambda)$,we get $x^{2} + y^{2} - \frac{6}{1+\lambda}x - \frac{4\lambda}{1+\lambda}y = 0$.The centre of this circle is $\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}ight)$.Since the centre lies on the line $2x - 3y + 12 = 0$,we substitute the coordinates:$2\left(\frac{3}{1+\lambda}ight) - 3\left(\frac{2\lambda}{1+\lambda}ight) + 12 = 0$$6 - 6\lambda + 12(1+\lambda) = 0$$6 - 6\lambda + 12 + 12\lambda = 0$$6\lambda = -18 \Rightarrow \lambda = -3$.Substituting $\lambda = -3$ into the equation of the circle:$(x^{2}+y^{2}-6x) - 3(x^{2}+y^{2}-4y) = 0$$-2x^{2} - 2y^{2} - 6x + 12y = 0$$x^{2} + y^{2} + 3x - 6y = 0$.Checking the point $(-3, 6)$ in the equation:$(-3)^{2} + (6)^{2} + 3(-3) - 6(6) = 9 + 36 - 9 - 36 = 0$.Thus,the circle passes through the point $(-3, 6)$.

The circle passing through the intersection of the circles $x^{2}+y^{2}-6x=0$ and $x^{2}+y^{2}-4y=0$,having its centre on the line $2x-3y+12=0$,also passes through the point:

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