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

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

## 1 Answer

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

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer