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The equation of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the x- and y- axis respectively, are :
\[\begin {array} {1 1} (1)\;x^2+y^2 \pm 4x \pm 8y=0 & \quad (2)\;x^2+y^2 \pm 2x \pm 4y=0 \\ (3)\;x^2+y^2\pm 8x \pm 16 y=0 & \quad (4)\;x^2+y^2 \pm 8x \pm 16 y=0 \end {array}\]
jeemain
eamcet
math
2009
q49
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asked
Oct 10, 2013
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meena.p
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$(1)\;x^2+y^2 \pm 4x \pm 8y=0$
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Nov 7, 2013
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pady_1
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