Q. If the circles $x^{2} + y^{2} + 2x + 2ky + 11 = 6$ and $x^{2} + y^{2} + 2ky + K = 0$ intersect orthogonally, then $k$ is
A) $2\ or - \frac{3}{2}$
B) $- 2\ or\frac{3}{2}$
C) $2\ or\frac{3}{2}$
D) $- 2\ \frac{3}{2}$
Explanation
(a)
Given equation of circles are $x^{2} + y^{2} + 2x + 2ky + 6 = 0$ and$x^{2} + y^{2} + 2ky + k = 0$. They intersect each other orthogonally
$\therefore 2 \bullet 1 \bullet 0 + 2 \bullet k \bullet k = 6 + k$
$\Rightarrow \ 2k^{2} - k - 6 = 0$
$\Rightarrow \ (2k + 3)(k - 2) = 0$
$\Rightarrow \ k = 2,\ - \frac{3}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.