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(A) Given circles are $S_1 \equiv x^2+y^2+2x+2y+1=0$ and $S_2 \equiv x^2+y^2+4x+6y+4=0$. The equation of the common chord is $S_1 - S_2 = 0$: $(x^2+y^

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$10x^2+10y^2+14x+8y+1=0$

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(A) Given circles are $S_1 \equiv x^2+y^2+2x+2y+1=0$ and $S_2 \equiv x^2+y^2+4x+6y+4=0$. The equation of the common chord is $S_1 - S_2 = 0$: $(x^2+y^2+2x+2y+1) - (x^2+y^2+4x+6y+4) = 0$ $-2x - 4y - 3 = 0 \Rightarrow 2x + 4y + 3 = 0$. The equation of a circle passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$ (or $S_1 + \lambda(L) = 0$ where $L$ is the common chord). $x^2+y^2+2x+2y+1 + \lambda(2x+4y+3) = 0$ $x^2+y^2+2x(1+\lambda) + 2y(1+2\lambda) + (1+3\lambda) = 0$. The center of this circle is $(- (1+\lambda), -(1+2\lambda))$. Since the common chord $2x+4y+3=0$ is a diameter,the center must lie on it: $2(-(1+\lambda)) + 4(-(1+2\lambda)) + 3 = 0$ $-2 - 2\lambda - 4 - 8\lambda + 3 = 0$ $-10\lambda - 3 = 0 \Rightarrow \lambda = -\frac{3}{10}$. Substituting $\lambda = -\frac{3}{10}$ into the circle equation: $x^2+y^2+2x(1-\frac{3}{10}) + 2y(1-\frac{6}{10}) + (1-\frac{9}{10}) = 0$ $x^2+y^2+\frac{14x}{10} + \frac{8y}{10} + \frac{1}{10} = 0$ $10x^2+10y^2+14x+8y+1=0$.

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2+4x+6y+4=0$ is

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