BPSC 46th Exam Prelims 2004 Solved Paper

From the question,
${\mathit{x}}^2+{\mathit{y}}^2-6\mathit{x}-2\mathit{y}-3=0$
After comparing it with general equation of circle
${\mathit{x}}^2+{\mathit{y}}^2+2\mathit{g}\mathit{x}+2\mathit{f}\mathit{y}+\mathit{c}=0$
$2\mathit{g}=-6\therefore \mathit{g}=-3$
$2\mathit{f}=-2\Rightarrow \mathit{f}=-1$
Now, the centre of th circle $=(-\mathit{g},$ $-\mathit{f})=\left(3,1\right)$
$\therefore$ Radius of the circle, which touches the line $\mathit{x}-\mathit{y}=0$ and concentric with circle ${\mathit{x}}^2+{\mathit{y}}^2-6\mathit{x}-2\mathit{y}-3=0$ will be
$=\frac{\left|3-1\right|}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$