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(D) Let the given circles be $S_1: 2x^2+2y^2-2x+6y-3=0$ and $S_2: x^2+y^2+4x+2y+1=0$.Normalize $S_1$ to $x^2+y^2-x+3y-\frac{3}{2}=0$.The equation of t

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$4x^2+4y^2+6x+10y-1=0$

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(D) Let the given circles be $S_1: 2x^2+2y^2-2x+6y-3=0$ and $S_2: x^2+y^2+4x+2y+1=0$.Normalize $S_1$ to $x^2+y^2-x+3y-\frac{3}{2}=0$.The equation of the common chord is $S_1 - 2S_2 = 0$:$(2x^2+2y^2-2x+6y-3) - 2(x^2+y^2+4x+2y+1) = 0$$-10x + 2y - 5 = 0 \Rightarrow 10x - 2y + 5 = 0$.The family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda S_2 = 0$:$(2+\lambda)x^2 + (2+\lambda)y^2 + (4\lambda-2)x + (2\lambda+6)y + (\lambda-3) = 0$.Dividing by $(2+\lambda)$,the centre is $(h, k) = \left(-\frac{4\lambda-2}{2(2+\lambda)}, -\frac{2\lambda+6}{2(2+\lambda)}\right) = \left(-\frac{2\lambda-1}{\lambda+2}, -\frac{\lambda+3}{\lambda+2}\right)$.Since the centre lies on the common chord $10x - 2y + 5 = 0$:$10\left(-\frac{2\lambda-1}{\lambda+2}\right) - 2\left(-\frac{\lambda+3}{\lambda+2}\right) + 5 = 0$.$-20\lambda + 10 + 2\lambda + 6 + 5\lambda + 10 = 0$ $\Rightarrow -13\lambda + 26 = 0$ $\Rightarrow \lambda = 2$.Substituting $\lambda = 2$ into the family equation: $(2+2)x^2 + (2+2)y^2 + (8-2)x + (6+4)y + (2-3) = 0$.Resulting in $4x^2+4y^2+6x+10y-1=0$.

The equation of a circle which passes through the points of intersection of the circles $2x^2+2y^2-2x+6y-3=0$ and $x^2+y^2+4x+2y+1=0$,and whose centre lies on the common chord of these circles is

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