The condition for the coaxial system $x^{2} + y^{2} + 2\lambda x + c = 0$, where $\lambda$ is a parameter and $c$ is a constant to have distinct limiting points, is

Question

Dr. Abubacker · Accepted Answer

(d)
Centre and radius of given circle are $( - \lambda,\ 0)$ and
$r = \sqrt{\lambda^{2} - c}$
For limiting, point $r = 0,\ \lambda = \pm \sqrt{c}$
Thus, we get two limiting points of the given coaxial system as $( \pm \sqrt{c},\ 0)$
For real and distinct $c > 0$

Q. The condition for the coaxial system $x^{2} + y^{2} + 2\lambda x + c = 0$, where $\lambda$ is a parameter and $c$ is a constant to have distinct limiting points, is

Explanation

Reviewed by Dr. Abubacker

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