Q. If $x - y + 1 = 0$ meets the circle $x^{2} + y^{2} + y - 1 = 0$ at $A$ and $B$, then the equation of the circle with $AB$ as diameter is
A) $2\left( x^{2} + y^{2} \right) + 3x - y + 1 = 0$
B) $2\left( x^{2} + y^{2} \right) + 3x - y + 2 = 0$
C) $2\left( x^{2} + y^{2} \right) + 3x - y + 3 = 0$
D) $x^{2} + y^{2} + 3x - y + 1 = 0$
Explanation
(a)
The intersection points of line and circle are $A\left( - \frac{1}{2},\frac{1}{2} \right)$ and $B( - 1,\ 0)$
These are the end points of a diameter
$\therefore$ The equation of circle is
$\left( x + \frac{1}{2} \right)(x + 1) + \left( y - \frac{1}{2} \right)(y - 0) = 0$
$\Rightarrow \ \ \ (2x + 1)(x + 1) + (2y - 1)y = 0$
$\Rightarrow \ \ 2\left( x^{2} + y^{2} \right) + 3x - y + 1 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.