Q. The equation of circle which touches the $x$-axis and $y$-axis at the points (1, 0) and (0, 1) respectively, is
A) $x^{2} + y^{2} - 4y + 3 = 0$
B) $x^{2} + y^{2} - 2y - 2 = 0$
C) $x^{2} + y^{2} - 2x - 2y + 2 = 0$
D) $x^{2} + y^{2} - 2x - 2y + 1 = 0$
Explanation
(d)
Since circle touches the $x$-axis and$\ y$-axis at points (1, 0) respectively.So, centre of circle is (1, 1)and radius is 1
[image:11_image11.emf]
Hence, equation of circle is
$(x - 1)^{2} + (y - 1)^{2} = 1^{2}$
$\Rightarrow \ \ x^{2} + y^{2} - 2x - 2y + 1 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.