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If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0,$ then $c=$

\[(a)\;16\quad (b)\;24 \quad (c)\;-42 \quad (d)\;-62 \]

1 Answer

(d) -62
answered Nov 7, 2013 by pady_1
 
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