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. $x^2+y^2+3x-y+4=0$

Answer 1

Answer: (A)

Equation of the circle passing through intersecting point of line and circle is given by $S+\lambda L=0$
where, $S{x}^2+y^2+y-1=0$
and $Lx-y+1=0$
$\therefore$ Equation of the circle is
$\left(x^2+y^2+y-1\right)+\lambda (x-y+1)=0$
$\Rightarrow x^2+y^2+\lambda x+(1-\lambda )y+\lambda -1=0$
Centre of above circle $=\left(\frac{-\lambda }{2},\frac{\lambda -1}{2}\right)$
Since, centre lies on $x-y+1=0$.
$\therefore \frac{-\lambda }{2}-\left(\frac{\lambda -1}{2}\right)+1=0\Rightarrow \lambda =\frac{3}{2}$
Hence, required equation of circle is
$x^2+y^2+y-1+\frac{3}{2}(x-y+1)=0$
$\Rightarrow 2\left(x^2+y^2\right)+3x-y+1=0$