Q. Which of the following is a point on the common chord of the circles $x^{2} + y^{2} + 2x - 3y + 6 = 0$ and $x^{2} + y^{2} + x - 8y - 13 = 0$
A) (1, 4)
B) $(1, - 2)$
C) $(1, - 4)$
D) (1, 2)
Explanation
(c)
Let $S_{1} = x^{2} + y^{2} + 2x - 3y + 6 = 0\$
$S_{2} = x^{2} + y^{2} + x - 8y - 13 = 0\$
So, the common chord is given by
$S_{1} - S_{2} = 0$
$\therefore$ Common chord is
$x + 5y + 19 = 0$
and this equation of common chord is satisfied by $(1, - 4\ )$
only
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.