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(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The centre is $(-\alpha, -\beta)$ which is given as $(\alpha, \beta)$,so $g=-\alpha$ and

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$-3$

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(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The centre is $(-\alpha, -\beta)$ which is given as $(\alpha, \beta)$,so $g=-\alpha$ and $f=-\beta$. The circle $x^2+y^2+2gx+2fy+c=0$ cuts $x^2+y^2-2y-3=0$ orthogonally. The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here $g_1=g, f_1=f, c_1=c$ and $g_2=0, f_2=-1, c_2=-3$. So,$2g(0) + 2f(-1) = c-3 \implies -2f = c-3 \implies c = 3-2f = 3+2\beta$. Next,it cuts $x^2+y^2+4x+3=0$ orthogonally. Here $g_3=2, f_3=0, c_3=3$. So,$2g(2) + 2f(0) = c+3 \implies 4g = c+3 \implies 4(-\alpha) = c+3 \implies c = -4\alpha-3$. Equating the two expressions for $c$: $3+2\beta = -4\alpha-3 \implies 4\alpha+2\beta = -6 \implies 2\alpha+\beta = -3$. Since the centre $(\alpha, \beta)$ lies on $2x-3y+4=0$,we have $2\alpha-3\beta+4=0$. From $2\alpha+\beta = -3$,we get $2\alpha = -3-\beta$. Substituting this into the line equation: $(-3-\beta)-3\beta+4=0 \implies -4\beta+1=0 \implies \beta=1/4$. Then $2\alpha = -3-1/4 = -13/4 \implies \alpha = -13/8$. The value $2\alpha+\beta = -3$.

If the centre $(\alpha, \beta)$ of a circle cutting the circles $x^2+y^2-2y-3=0$ and $x^2+y^2+4x+3=0$ orthogonally lies on the line $2x-3y+4=0$,then $2\alpha+\beta=$

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