The equation of the circle,the end-points of | vedclass

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(A) The center of the first circle $x^{2}+y^{2}-2x+3y-3=0$ is found by comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,giving $C_{1} = (-g, -f) = (1, -3/2)$.

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$2x^{2}+2y^{2}+4x-9y-24=0$

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(A) The center of the first circle $x^{2}+y^{2}-2x+3y-3=0$ is found by comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,giving $C_{1} = (-g, -f) = (1, -3/2)$.The center of the second circle $x^{2}+y^{2}+6x-12y-5=0$ is $C_{2} = (-3, 6)$.The equation of a circle with diameter endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) = 0$.Substituting the centers $(1, -3/2)$ and $(-3, 6)$:$(x-1)(x+3) + (y+3/2)(y-6) = 0$$x^{2}+2x-3 + y^{2}-6y+3/2y-9 = 0$$x^{2}+y^{2}+2x-9/2y-12 = 0$Multiplying by $2$ to match the options:$2x^{2}+2y^{2}+4x-9y-24 = 0$.

The equation of the circle,the end-points of whose diameter are the centres of the circles $x^{2}+y^{2}-2x+3y-3=0$ and $x^{2}+y^{2}+6x-12y-5=0$ is

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