Q. The value of $\lambda$, for which the circle $x^{2} + y^{2} + 2\lambda x + 6y + 1 = 0$ intersects the circle $x^{2} + y^{2} + 4x + 2y = 0$ orthogonally, is
A) $\frac{11}{8}$
B) $- 1$
C) $\frac{- 5}{4}$
D) $\frac{5}{2}$
Explanation
(c)
Here, $g_{1} = \lambda,\ \ f_{1} = 3,\ \ c_{1} = 1$
and $g_{2} = 2,\ \ f_{2} = 1,\ \ c_{2} = 0$
since, they intersect orthogonally
$\therefore\ \ \ 2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$
$\Rightarrow \ \ 2\lambda \times 2 + 6 \times 1 = 1 + 0$
$\Rightarrow \ \ 4\lambda + 6 = 1$
$\Rightarrow \ \ \lambda = - \frac{5}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.