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Home  >>  EAMCET  >>  Mathematics
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The inverse point of (1,2) with respect to the circle $x^2+y^2-4x-6y+9=0$ is

\[\begin {array} {1 1} (a)\;(0,0) & \quad (b)\;(1,0) \\ (c)\;(0,1) & \quad (d)\;(1,1) \end {array}\]
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(c) (0,1)
answered Nov 7, 2013 by pady_1
 
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