Q. The locus of centres of family of circle passing through the origin and cutting the circle $x^{2} + y^{2} + 4x - 6y - 13 = 0$ orthogonally, is
A) $4x + 6y + 13 = 0$
B) $4x - 6y + 13 = 0$
C) $4x + 6y - 13 = 0$
D) $4x - 6y - 13 = 0$
Explanation
(d)
Let the family of circles passing through origin be
$x^{2} + y^{2} + 2gx + 2fy = 0$
They intersect circle $x^{2} + y^{2} + 4x - 6y - 13 = 0$
Orthogonally
So, $2g(2) - 2f(3) = - 13$
Hence, locus of $( - g,\ - f)$ is
$- 4x + 6y + 13 = 0$
$\Rightarrow 4x - 6y - 13 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.