NIMCET 2013 Question Paper with solutions

Since, centre lies on the line $\mathit{x}+2\mathit{y}+3=0$
Let the centre be (- 2a - 3 a), then its distance from the given two points (-1,1) and (2,1) is
$\Rightarrow \mathit{a}=-\frac{7}{4}$
Thus, the centre is $\left(\frac{1}{2},\frac{7}{4}\right)$
Radius is $\sqrt{\frac{157}{16}}$
So, the equation of the circle is
$\mathit{x}{\left(\mathit{x}-\frac{1}{2}\right)}^2+{\left(\mathit{y}+\frac{7}{4}\right)}^2=\frac{157}{16}$
$\Rightarrow 2{\mathit{x}}^2+2{\mathit{y}}^2-2\mathit{x}+7\mathit{y}-13=0$