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

$\begin {array} {1 1} (1)\;\bigg(1,\frac{1}{2}\bigg) & \quad (2)\;(2,1) \\ (3)\;(0,1) & \quad (4)\;(1,0) \end {array}$
Can you answer this question?

(3) (0,1)
answered Nov 7, 2013 by