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The diameter of a circle are along $2x+y-7=0$ and $x+3y-11=0$. Then, the equation of this circle, Which also passes through (5,7) is :

\[\begin {array} {1 1} (1)\;x^2+y^2-4x-6y-16=0 & \quad (2)\;x^2+y^2-4x-6y-20=0 \\ (3)\;x^2+y^2-4x-6y-12=0 & \quad (4)\;x^2+y^2+4x+6y-12=0 \end {array}\]

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$ (3)\;x^2+y^2-4x-6y-12=0$
answered Nov 7, 2013 by pady_1

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