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. $2(x^2+y^2+3x-y+1=0$
• B. $2(x^2+y^2)+3x-y+2=0$
• C. $2(x^2+y^2)+3x-y+3=0$
• D. $x^2+y^2+3x-y+1=0$

Answer 1

The correct option is A $2(x^2+y^2+3x-y+1=0$

$S+\lambda L=0$, $\lambda$ being a constant.
$x^2+y^2+y-1+\lambda (x-y+1)=0$
Center $=(\frac{-\lambda }{2},\frac{\lambda -1}{2})$
Centre lies on the line $x-y+1=0$
$\Rightarrow \lambda =\frac{3}{2}$
Equation thus becomes $x^2+y^2+y-1+\frac{3}{2}(x-y+1)=0$
i.e. $2(x^2+y^2)+3x-y+1=0$