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Mathematics
The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6x+8=0$ and $x^2+y^2-2x-2y=7$ is :
\[\begin {array} {1 1} (1)\;3x^2+3y^2-8x-13 y=0 & \quad (2)\;3x^2+3y^2-8x+29y=0 \\ (3)\;3x^2+3y^2+8x+29 y=0 & \quad (4)\;3x^2+3y^2-8x-29 y=0 \end {array}\]
jeemain
eamcet
math
2009
q53
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asked
Oct 10, 2013
by
meena.p
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$(2)\;3x^2+3y^2-8x+29y=0$
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Nov 7, 2013
by
pady_1
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