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

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

1 Answer

The given lines can be re-written as
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)
Hence ABCD be a square and AC and BD are the diameters of the circle .
After solving we get,
Equation of circle is $(x-0)(x+1)+(y-0)(y+1)=0$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v

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