NIMCET 2015 Question Paper with solutions

Equation of first circle ${\mathit{x}}^2+{\mathit{y}}^2+2\mathit{g}\mathit{x}+2\mathit{f}\mathit{y}=0$ can be written as ${\left(\mathit{x}+\mathit{g}\right)}^2+{\left(\mathit{y}+\mathit{f}\right)}^2={\mathit{g}}^2+{\mathit{f}}^2$
Center of first circle $\left(-\mathit{g},-\mathit{f}\right)$ and radius $\sqrt{{\mathit{g}}^2+{\mathit{f}}^2}$
Equation of second circle ${\mathit{x}}^2+{\mathit{y}}^2+2{\mathit{g}}^{\prime }\mathit{s}+2{\mathit{f}}^{\prime }\mathit{y}=0$ can be written as ${\left(\mathit{x}+{\mathit{g}}^{\prime }\right)}^2+{\left(\mathit{y}+{\mathit{f}}^{\prime }\right)}^2={\mathit{g}}^{\prime 2}+{\mathit{f}}^{\prime 2}$
Center of first circle $\left(-{\mathit{g}}^{\prime },-{\mathit{f}}^{\prime }\right)$ and radius $\sqrt{{\mathit{g}}^{\prime 2}+{\mathit{f}}^{\prime 2}}$
The two circle touch each other so $\sqrt{{\left(-\mathit{g}+{\mathit{g}}^{\prime }\right)}^2+{\left(-\mathit{f}+{\mathit{f}}^{\prime }\right)}^2}=\sqrt{{\mathit{g}}^2+{\mathit{f}}^2}-\sqrt{{\mathit{g}}^{\prime 2}+{\mathit{f}}^{\prime 2}}$
By squaring on both sides we get
${\left(-\mathit{g}+{\mathit{g}}^{\prime }\right)}^2+{\left(-\mathit{f}+{\mathit{f}}^{\prime }\right)}^2={\left(\sqrt{{\mathit{g}}^2+{\mathit{f}}^2}-\sqrt{{\mathit{g}}^{\prime 2}+{\mathit{f}}^{\prime 2}}\right)}^2$
$\mathit{g}{\mathit{g}}^{\prime }+\mathit{f}{\mathit{f}}^{\prime }=\sqrt{{\mathit{g}}^2+{\mathit{f}}^2}\sqrt{{\mathit{g}}^{\prime 2}+{\mathit{f}}^{\prime 2}}$
Again by squaring on both sides we get
${\left(\mathit{g}{\mathit{g}}^{\prime }+\mathit{f}{\mathit{f}}^{\prime }\right)}^2={\left(\sqrt{{\mathit{g}}^2+{\mathit{f}}^2}\sqrt{{\mathit{g}}^{\prime 2}+{\mathit{f}}^{\prime 2}}\right)}^2$
$2\mathit{g}{\mathit{g}}^{\prime }\mathit{f}{\mathit{f}}^{\prime }=\left(\mathit{g}{\mathit{f}}^{\prime 2}\right)+{\left({\mathit{g}}^{\prime }\mathit{f}\right)}^2$
$\left(\mathit{g}{\mathit{f}}^{\prime 2}\right)+{\left({\mathit{g}}^{\prime }\mathit{f}\right)}^2-2\mathit{g}{\mathit{g}}^{\prime }\mathit{f}{\mathit{f}}^{\prime }=0$
${\left(\mathit{g}{\mathit{f}}^{\prime }-{\mathit{g}}^{\prime }\mathit{f}\right)}^2=0\Rightarrow \mathit{g}{\mathit{f}}^{\prime }-{\mathit{g}}^{\prime }\mathit{f}=0$
$\mathit{g}{\mathit{f}}^{\prime }=\mathit{g}$ 'f