Q. The equation of the circle touching $x = 0,\ y = 0$ and $x = 4$ is
A) $x^{2} + y^{2} - 4x - 4y + 16 = 0$
B) $x^{2} + y^{2} - 8x - 8y + 16 = 0$
C) $x^{2} + y^{2} + 4x + 4y - 4 = 0$
D) $x^{2} + y^{2} - 4x - 4y + 4 = 0$
Explanation
(d)
Since, the required circle touch $x = 0,\ y = 0$ and $x = 4$
Centre is (2, 2) and radius=2
$\therefore$ Required circle is
$(x - 2)^{2} + (y - 2)^{2} = (2)^{2}$
$\Rightarrow x^{2} + y^{2} - 4x - 4y + 4 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.