NIMCET 2016 Question Paper with solutions

Given, equation of circles are
$2\mathit{x}-3\mathit{y}+12=0$...(i)
$\mathit{x}+4\mathit{y}-5=0$....(ii)
On solving Eq (i) and Eq (ii). we get
$\mathit{x}=-3$ and y$=2$
Hence coordinate of centre is$\left(-3,2\right)$
Given area of circle = 154 sq units
$\Rightarrow \mathit{\pi }{\mathit{r}}^2=154$   [where r is radius of circle]
$\Rightarrow \frac{22}{7}\times {\mathit{r}}^2=154$
${\mathit{r}}^2=49$
$\Rightarrow \mathit{r}=7$
Here centre is $\left(-32\right)$ and radius $=7$
Then, required of equation of circle is
${\left\{\mathit{x}-\left(-3\right)\right\}}^2+{\left(\mathit{y}-2\right)}^2={\left(7\right)}^2$
$\Rightarrow {\left(\mathit{x}+3\right)}^2+{\left(\mathit{y}-2\right)}^2=49$
$\Rightarrow {\mathit{x}}^2+9+6\mathit{x}+{\mathit{y}}^2+4-4\mathit{y}-49$
$\Rightarrow {\mathit{x}}^2+{\mathit{y}}^2+6\mathit{x}-4\mathit{y}+13-49=0$
$\Rightarrow {\mathit{x}}^2+{\mathit{y}}^2+6\mathit{x}-4\mathit{y}-36=0$