Q. The circle $x^{2} + y^{2} + 8y - 4 = 0$, cuts the real circle $x^{2} + y^{2} + gx + 4 = 0$ orthogonally, if
A) For any real value of $g$
B) For no real value of $g$
C) $g = 0$
D) $g < - 2,\ g > 2$
Explanation
(a)
The given circles cuts orthogonally, if
$2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$
$\therefore\ \ \ 2 \times \frac{g}{2} \times 0 + 2 \times 4 \times 0 = 4 - 4$
This is true for any real value of $g$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.