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(C) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2+2x+4y+1=0$ and $S_2: x^2+y^2-2x-4y-4=0$ is given by $S_1 +

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$\sqrt{39}$

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(C) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2+2x+4y+1=0$ and $S_2: x^2+y^2-2x-4y-4=0$ is given by $S_1 + \lambda S_2 = 0$. $(1+\lambda)x^2 + (1+\lambda)y^2 + (2-2\lambda)x + (4-4\lambda)y + (1-4\lambda) = 0$. Dividing by $(1+\lambda)$,we get $x^2+y^2 + \frac{2(1-\lambda)}{1+\lambda}x + \frac{4(1-\lambda)}{1+\lambda}y + \frac{1-4\lambda}{1+\lambda} = 0$. This circle intersects $x^2+y^2-6=0$ orthogonally. The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here $g_1 = \frac{1-\lambda}{1+\lambda}, f_1 = \frac{2(1-\lambda)}{1+\lambda}, c_1 = \frac{1-4\lambda}{1+\lambda}$ and $g_2=0, f_2=0, c_2=-6$. Substituting these,$2(\frac{1-\lambda}{1+\lambda})(0) + 2(\frac{2(1-\lambda)}{1+\lambda})(0) = \frac{1-4\lambda}{1+\lambda} - 6$. $0 = \frac{1-4\lambda - 6 - 6\lambda}{1+\lambda} \implies 1-4\lambda-6-6\lambda = 0 \implies -10\lambda = 5 \implies \lambda = -1/2$. Substituting $\lambda = -1/2$ into the circle equation: $(1-1/2)x^2 + (1-1/2)y^2 + (2+1)x + (4+2)y + (1+2) = 0$. $0.5x^2 + 0.5y^2 + 3x + 6y + 3 = 0 \implies x^2+y^2+6x+12y+6=0$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{3^2+6^2-6} = \sqrt{9+36-6} = \sqrt{39}$.

The radius of the circle passing through the points of intersection of the circles $x^2+y^2+2x+4y+1=0$ and $x^2+y^2-2x-4y-4=0$ and intersecting the circle $x^2+y^2=6$ orthogonally is

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