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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 $A B$ as diameter is
EAMCETEAMCET 2005
Solution:
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)]
$\Rightarrow 2 x^{2}+3 x+1=0$
$\Rightarrow (2 x+1)(x+1)=0$
$\Rightarrow x=-\frac{1}{2},-1$ 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
$\left(x+\frac{1}{2}\right)(x+1)+\left(y-\frac{1}{2}\right)(y-0)=0$
$\Rightarrow (2 x+1)(x+1)+(2 y-1) y=0$
$\Rightarrow 2 x^{2}+x+2 x+1+2 y^{2}-y=0$
$\Rightarrow 2\left(x^{2}+y^{2}\right)+3 x-y+1=0$