$\begin{array}{1 1}(a)\;x^2+x+y^2+y=0\\(b)\;2x^2+x+2y^2+y=0\\(c)\;x^2-x+y^2-y=0\\(d)\;x^2-x-y^2-y=0\end{array}$

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The given lines can be re-written as

$x+y+1=0$------(1)

$x+y-1=0$------(2)

$x-y+1=0$------(3)

$x-y-1=0$------(4)

Equation (1) and (2) are parallel and equation (3) and (4) are also parallel.

Slope of (1) or (2) =$-1=m_1$(say)

Slope of (3) or (4) =$1=m_2$(say)

$m_1m_2=-1$

Hence ABCD be a square and AC and BD are the diameters of the circle .

After solving we get,

$A=(0,-1)$

$C=(-1,0)$

Equation of circle is $(x-0)(x+1)+(y-0)(y+1)=0$

$x^2+x+y^2+y=0$

Hence (a) is the correct answer.

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