Q. The equation of the circle whose diameter is the common chord of the circles $x^{2} + y^{2} + 2x + 3y + 2 = 0$ and $x^{2} + y^{2} + 2x - 3y - 4 = 0$ is
A) $x^{2} + y^{2} + 2x + 2y + 2 = 0$
B) $x^{2} + y^{2} + 2x + 2y - 1 = 0$
C) $x^{2} + y^{2} + 2x + 2y + 1 = 0$
D) $x^{2} + y^{2} + 2x + 2y + 3 = 0$
Explanation
(c)
Equation of common chord is
$S_{1} - S_{2} = 0$
$\Rightarrow \ \ \ 6y + 6 = 0$
$\Rightarrow \ \ \ y = - 1$
On putting $y = - 1$ in first circle,
$\therefore\ \ \ x^{2} + 1 + 2x - 3 + 2 = 0$
$\Rightarrow \ \ x^{2} + 2x = 0\ \ \Rightarrow \ \ x = 0,\ - 2$
$\therefore$ End points of diameter are $(0,\ - 1)$ and $( - 2,\ - 1)$
Equation of circle is
$(x - 0)(x + 2) + (y + 1)(y + 1) = 0$
$\Rightarrow \ \ x^{2} + 2x + y^{2} + 2y + 1 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.