Q. The limiting points of the coaxial system of circles $x^{2} + y^{2} + 2\lambda y + 4 = 0$ are
A) $(0,\ \pm 4)$
B) $( \pm 2,\ 0)$
C) $(0,\ \pm 1)$
D) $(0,\ \pm 2)$
Explanation
(d)
The centre of given circle is $(0,\ - \lambda)$
$\therefore\ \ \ r = \sqrt{0 + \lambda^{2} - 4} = 0$
$\Rightarrow \ \ \ \ \lambda = \pm 2$
So, limiting points are $(0,\ \pm 2)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.