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) $(1,\ - 2)$
B) $(1,\ 4)$
C) $(1,\ 2)$
D) $(1,\ - 4)$
Explanation
(d)
Let the equation of circles are
$S_{1} \equiv x^{2} + y^{2} + 2x - 3y + 6 = 0$ …(i)
$S_{2} \equiv x^{2} + y^{2} + x - 8y - 13 = 0$ …(ii)
$\therefore$ Equation of common chord is
$S_{1} - S_{2} = 0$
$\Rightarrow \left( x^{2} + y^{2} + 2x - 3y + 6 \right) - \left( x^{2} + y^{2} + x - 8y - 13 \right) = 0$
$\Rightarrow \ \ x + 5y + 19 = 0$ …(iii)
In the given option only the point $(1,\ - 4)$ satisfied the Eq. (iii)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.