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(A) The family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$. Substituting the gi

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$x^2+y^2+10x+3y+5=0$

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(A) The family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$. Substituting the given equations: $(x^2+y^2+2x+3y+1) + \lambda(x^2+y^2+4x+3y+2) = 0$. Since the circle passes through the point $(-1, 1)$,we substitute $x = -1$ and $y = 1$: $((-1)^2 + (1)^2 + 2(-1) + 3(1) + 1) + \lambda((-1)^2 + (1)^2 + 4(-1) + 3(1) + 2) = 0$. $(1 + 1 - 2 + 3 + 1) + \lambda(1 + 1 - 4 + 3 + 2) = 0$. $4 + 3\lambda = 0 \Rightarrow \lambda = -\frac{4}{3}$. Substituting $\lambda = -\frac{4}{3}$ back into the family equation: $(x^2+y^2+2x+3y+1) - \frac{4}{3}(x^2+y^2+4x+3y+2) = 0$. $3(x^2+y^2+2x+3y+1) - 4(x^2+y^2+4x+3y+2) = 0$. $3x^2+3y^2+6x+9y+3 - 4x^2-4y^2-16x-12y-8 = 0$. $-x^2-y^2-10x-3y-5 = 0$. $x^2+y^2+10x+3y+5 = 0$.

The equation of the circle passing through the points of intersection of two circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ and the point $(-1,1)$ is

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