# Find the equation of the circumcircle of the quadilateral formed by the four lines $x+y\pm1=0$ and $x-y\pm 1=0$?

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

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.