Q. The equation of the circle of radius 5 and touching the coordinate axes in third quadrant is
A) $({x - 5)}^{2} + (y + 5)^{2} = 25$
B) $({x + 4)}^{2} + (y + 4)^{2} = 25$
C) $({x + 6)}^{2} + (y + 6)^{2} = 25$
D) $({x + 5)}^{2} + (y + 5)^{2} = 25$
Explanation
(d)
Let $( - h,\ - k)$ be the centre of the circle
[image:11_image35.emf]
Circle touches the coordinate axes in IIIrd quadrant
$\therefore Radius = - h = - k$
$\Rightarrow h = k = - 5$
$\therefore$ The required equation of circle is
$(x + 5)^{2} + (y + 5)^{2} = 25$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.