Q. If the circles $x^{2} + y^{2} + 2ky + 2x + 6 = 0$ and $x^{2} + y^{2} + 2ky + k = 0$ intersects orthogonally. Then, the value of $k$ is
A) $\frac{3}{2}$
B) $- 2$
C) $- \frac{3}{2}$
D) $\frac{1}{2}$
Explanation
(c)
Here, $g_{1} = 1,\ \ f_{1} = k,\ \ c_{1} = 6$
and $g_{2} = 0,\ \ f_{2} = k,\ \ \ c_{2} = k$
Since, circles intersects orthogonally
$\therefore\ \ \ 2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$
$\Rightarrow \ \ \ \ 0 + 2k^{2} = 6 + k$
$\Rightarrow \ \ 2k^{2} - k - 6 = 0$
$\Rightarrow \ \ k = 2,\ - \frac{3}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.