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(A) Let the required circle be $S_2 = 0$. The point of contact is $P(1,2)$. The center of the given circle $S_1: x^2+y^2+2x+8y-23=0$ is $C_1(-1,-4)$ a

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$5$

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(A) Let the required circle be $S_2 = 0$. The point of contact is $P(1,2)$. The center of the given circle $S_1: x^2+y^2+2x+8y-23=0$ is $C_1(-1,-4)$ and its radius $r_1 = \sqrt{(-1)^2+(-4)^2-(-23)} = \sqrt{1+16+23} = \sqrt{40} = 2\sqrt{10}$.Since the circles touch externally,the distance between centers $C_1C_2 = r_1 + r_2 = 2\sqrt{10} + \sqrt{10} = 3\sqrt{10}$.The center $C_2(h,k)$ lies on the line joining $C_1(-1,-4)$ and $P(1,2)$. The vector $\vec{C_1P} = (1-(-1), 2-(-4)) = (2,6)$.The unit vector along $\vec{C_1P}$ is $\frac{(2,6)}{\sqrt{2^2+6^2}} = \frac{(2,6)}{\sqrt{40}} = \frac{(2,6)}{2\sqrt{10}} = (\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}})$.Since $C_2$ is at a distance $r_2 = \sqrt{10}$ from $P(1,2)$ along the direction $\vec{C_1P}$,we have $C_2 = P + r_2 	imes (	ext{unit vector}) = (1,2) + \sqrt{10}(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}) = (1+1, 2+3) = (2,5)$.The equation of the circle with center $(2,5)$ and radius $\sqrt{10}$ is $(x-2)^2 + (y-5)^2 = (\sqrt{10})^2$.$x^2-4x+4 + y^2-10y+25 = 10$.$x^2+y^2-4x-10y+19 = 0$.Comparing with $x^2+y^2+ax+by+c=0$,we get $a=-4, b=-10, c=19$.$|a+b+c| = |-4-10+19| = |5| = 5$.

If the equation of the circle whose radius is $\sqrt{10}$ and which touches the circle $x^2+y^2+2x+8y-23=0$ externally at the point $(1,2)$ is $x^2+y^2+ax+by+c=0$,then $|a+b+c|=$

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