Q. The equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines $x = 0$ and $y = 0$ is
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) $x^{2} + y^{2} + 6x + 6y + 9 = 0$
Explanation
(a)
Since, the circle touching both the coordinates axes in fourth quadrant, so equation is
$(x - 3)^{2} + (y + 3)^{2} = 3^{2}$
$\Rightarrow \ x^{2} + y^{2} - 6x + 6y + 9 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.