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The equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines x=0 and y=0 is

\[\begin {array} {1 1} (a)\;x^2+y^2-6x+6y+9=0 \\ (b)\;x^2+y^2-6x-6y+9=0 \\ (c)\;x^2+y^2+6x-6y+9=0  \\ (d)\;x^2+y^2+6x+6y+9=0 \end {array}\]

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1 Answer

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$(a)\;x^2+y^2-6x+6y+9=0 $
answered Nov 7, 2013 by pady_1
 

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