Question

The equation of the circle of least radius belonging to the coaxial system of circles
orthogonal to the system $x^2+y^2+2\lambda x+4=0$ is

• A. $x^2+y^2=0$
• B. $x^2+y^2=4$
• C. $x^2+y^2+2x-4=0$
• D. $x^2+y^2+2\lambda x+4=0$

Answer 1

The correct option is B $x^2+y^2=4$

Centre of the given coaxial system is, $(\lambda ,0)$
and radius is $=\sqrt{{\lambda }^2-4}=0$ for limiting point.
$\lambda =\pm 2$. Thus limit points are $(-2,0),(2,0)$
We know equation of the circle of having least radius belonging to the coaxial system of circles orthogonal to the system is circle having diameter limiting points.
Hence it's equation is given by, $(x-2)(x+2)+y^2=0\Rightarrow x^2+y^2=4$