If the circle x^2+y^2+2x+3y+1=0 cuts another | vedclass

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(B) The equations of the circles are $S_1 \equiv x^2+y^2+2x+3y+1=0$ and $S_2 \equiv x^2+y^2+4x+3y+2=0$. Since the circles intersect at $A$ and $B$,the

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$2x^2+2y^2+2x+6y+1=0$

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(B) The equations of the circles are $S_1 \equiv x^2+y^2+2x+3y+1=0$ and $S_2 \equiv x^2+y^2+4x+3y+2=0$. Since the circles intersect at $A$ and $B$,the equation of the common chord $AB$ is given by $S_1 - S_2 = 0$. $(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$ $-2x - 1 = 0 \Rightarrow x = -\frac{1}{2}$. The family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$. However,the circle with $AB$ as diameter is given by $S_1 + k(S_1 - S_2) = 0$ where the coefficient of $x^2$ and $y^2$ is $1$. Alternatively,using the property that the circle with diameter $AB$ is $S_1 + k(S_1 - S_2) = 0$,we find the equation. Substituting $x = -\frac{1}{2}$ into $S_1$: $\frac{1}{4} + y^2 - 1 + 3y + 1 = 0 \Rightarrow y^2 + 3y + \frac{1}{4} = 0$. The circle with diameter $AB$ has the equation $S_1 + k(S_1 - S_2) = 0$. $x^2+y^2+2x+3y+1 + k(2x+1) = 0$. Since the center of this circle must lie on the line $x = -\frac{1}{2}$,we have $x + \frac{k}{1} = -\frac{1}{2}$ $\Rightarrow -1 + k = -\frac{1}{2}$ $\Rightarrow k = \frac{1}{2}$. Substituting $k = \frac{1}{2}$: $x^2+y^2+2x+3y+1 + \frac{1}{2}(2x+1) = 0$. $x^2+y^2+2x+3y+1 + x + \frac{1}{2} = 0 \Rightarrow x^2+y^2+3x+3y+\frac{3}{2} = 0$. Multiplying by $2$: $2x^2+2y^2+2x+6y+1 = 0$.

If the circle $x^2+y^2+2x+3y+1=0$ cuts another circle $x^2+y^2+4x+3y+2=0$ at points $A$ and $B$,then the equation of the circle with $AB$ as a diameter is

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