Question

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. $x^2+y^2+3x-y+1=0$

Answer 1

Answer: (A)

Given that
$x-y+1=0...(i)$
and $x^2+y^2+y-1=0...(ii)$
$\Rightarrow x^2+(x+1{)}^2+x+1-1=0$[from Eq. (i)]
$\Rightarrow 2x^2+3x+1=0$
$\Rightarrow (2x+1)(x+1)=0$
$\Rightarrow x=-\frac{1}{2},-1$ and $y=\frac{1}{2},0$
$\therefore$ Point of $A\left(-\frac{1}{2},\frac{1}{2}\right)$ and $B(-1,0)$
These are the end points of a diameter.
$\therefore$ The equation of circle is
$\left(x+\frac{1}{2}\right)(x+1)+\left(y-\frac{1}{2}\right)(y-0)=0$
$\Rightarrow (2x+1)(x+1)+(2y-1)y=0$
$\Rightarrow 2x^2+x+2x+1+2y^2-y=0$
$\Rightarrow 2\left(x^2+y^2\right)+3x-y+1=0$