NIMCET 2011 Question Paper with solutions

Given equation can be rewritten as $\frac{\mathit{x}}{3}+\frac{\mathit{Y}}{2}=1$ and $\frac{\mathit{x}}{2}+\frac{\mathit{y}}{3}=1$
Lines intercept the coordinate axes are
${\mathit{x}}^2+{\mathit{y}}^2+2\mathit{g}\mathit{x}+2\mathit{f}\mathit{y}+\mathit{c}=0$
Let equation of circle be
${\mathit{x}}^2+{\mathit{y}}^2+2\mathit{g}\mathit{x}+2\mathit{f}\mathit{y}+\mathit{c}=0$
Since, ail points are concyclic.
$\therefore 9+6\mathit{g}+\mathit{c}=0,4+4\mathit{f}+\mathit{c}=0$
$4+4\mathit{g}+\mathit{c}=0,9+6\mathit{f}+\mathit{c}=0$
On solving, we get
$\mathit{g}=-\frac{5}{2},\mathit{f}=-\frac{5}{2}$
$\therefore$ Centre of circle
$\left(-\mathit{g},-\mathit{f}\right)=\left(\frac{5}{2},\frac{5}{2}\right)$