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If the circle $x^2+y^2+2x+3y+1=0$ cuts another circle $x^2+y^2+4x+3y+2=0$ in A and B, then the equation of the circle with AB as a diameter is :

\[\begin {array} {1 1} (1)\;x^2+y^2+x+3y+3=0 & \quad (2)\; 2x^2+2y^2+2x+6y+1=0 \\ (3)\;x^2+y^2+x+6y+1=0 & \quad (4)\;2x^2+2y^2+x+3y+1=0 \end {array}\]
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$(2)\; 2x^2+2y^2+2x+6y+1=0$
answered Nov 7, 2013 by pady_1
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