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(A) The equations of the circles are $S_1: x^2+y^2+x-3y-10=0$ and $S_2: x^2+y^2+2x-y-20=0$. The common chord is given by $S_1 - S_2 = 0$,which is $(x^

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(A) The equations of the circles are $S_1: x^2+y^2+x-3y-10=0$ and $S_2: x^2+y^2+2x-y-20=0$. The common chord is given by $S_1 - S_2 = 0$,which is $(x^2+y^2+x-3y-10) - (x^2+y^2+2x-y-20) = 0$. This simplifies to $-x-2y+10=0$,or $x+2y-10=0$. The family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$. $(x^2+y^2+x-3y-10) + \lambda(x+2y-10) = 0$. $x^2+y^2+(1+\lambda)x+(-3+2\lambda)y+(-10-10\lambda) = 0$. The center of this circle is $(-\frac{1+\lambda}{2}, \frac{3-2\lambda}{2})$. Since the common chord $x+2y-10=0$ is the diameter,the center must lie on it: $-\frac{1+\lambda}{2} + 2(\frac{3-2\lambda}{2}) - 10 = 0$. $-1-\lambda + 6-4\lambda - 20 = 0 \implies -5\lambda - 15 = 0 \implies \lambda = -3$. Substituting $\lambda = -3$ into the circle equation: $x^2+y^2+(1-3)x+(-3-6)y+(-10+30) = 0$. $x^2+y^2-2x-9y+20=0$. Comparing with $x^2+y^2+\alpha x+\beta y+\gamma=0$,we get $\alpha=-2, \beta=-9, \gamma=20$. Thus,$\alpha+2\beta+\gamma = -2 + 2(-9) + 20 = -2 - 18 + 20 = 0$.

If the equation of the circle having the common chord of the circles $x^2+y^2+x-3y-10=0$ and $x^2+y^2+2x-y-20=0$ as its diameter is $x^2+y^2+\alpha x+\beta y+\gamma=0$,then $\alpha+2\beta+\gamma=$

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